All Exams >
EmSAT Achieve >
Physics for EmSAT Achieve >
All Questions
All questions of Centre of Mass for EmSAT Achieve Exam
The centre of mass of a body is locateda)outside the systemb)inside or outside the systemc)inside the systemd)at the centre of systemCorrect answer is option 'B'. Can you explain this answer?
|
Suresh Iyer answered |
The centre of mass of a body can lie within or outside the body.
Example
(i) Centre of mass of a uniform rod lies at its geometrical centre which lies within the rod
(ii) Centre of mass of a uniform ring lies at its geometrical centre which lies outside the ring.
Example
(i) Centre of mass of a uniform rod lies at its geometrical centre which lies within the rod
(ii) Centre of mass of a uniform ring lies at its geometrical centre which lies outside the ring.
Can you explain the answer of this question below:Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is
- A:
10 m/s
- B:
5 m/s
- C:
15 m/s
- D:
20 m/s
The answer is a.
Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is
10 m/s
5 m/s
15 m/s
20 m/s
|
EduRev JEE answered |
Velocity of heavier block (v2) = 14m/s
Velocity of lighter block (v1) = 0m/s
Velocity of centre of mass,
Velocity of lighter block (v1) = 0m/s
Velocity of centre of mass,
A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin
- a)Is zero
- b)Remains constant
- c)Goes on increasing
- d)Goes on decreasing
Correct answer is option 'B'. Can you explain this answer?
A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin
a)
Is zero
b)
Remains constant
c)
Goes on increasing
d)
Goes on decreasing
|
Krishna Iyer answered |
Angular momentum (L) is defined as the distance of the object from a rotation axis multiplied by the linear momentum
L = mv×y
L = mv×y
As the particle moves, m; v; and y, all remain unchanged at any point of time
⇒ L = constant
⇒ L = constant
There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are- a)( 2, -1 )
- b)( 8/3 ,-1/3 )
- c)( -1/3 , 8/3 )
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are
a)
( 2, -1 )
b)
( 8/3 ,-1/3 )
c)
( -1/3 , 8/3 )
d)
none of these
|
Riya Banerjee answered |
Body A has mass of 1kg and location (1,2)
Body B has mass of 2kg and location (-1,3)
Body B has mass of 2kg and location (-1,3)
mcxc = m1x1 + m2x2
(1+2) xc = (1 * 1) + (2 * -1)
xc = -1/3
(1+2) xc = (1 * 1) + (2 * -1)
xc = -1/3
Similarly,
mcyc = m1y1 + m2y2
(1 + 2) yc = (1 * 2) + (2 * 3)
yc= 8/3
mcyc = m1y1 + m2y2
(1 + 2) yc = (1 * 2) + (2 * 3)
yc= 8/3
Hence, the coordinates of the center of mass are (-1/3, 8/3).
The motion of a potter’s wheel is an example of
- a)rolling motion
- b)rotatory motion
- c)translatory motion
- d)precessional motion
Correct answer is option 'B'. Can you explain this answer?
The motion of a potter’s wheel is an example of
a)
rolling motion
b)
rotatory motion
c)
translatory motion
d)
precessional motion
|
Anjali Iyer answered |
Potter’s wheel is an example of rotary motion. Rotary motion is that kind of motion in which body of the mass moves along a circular path about an axis which remains fixed.
The centre of mass of a system of particles does not depend on
- a)masses of the particles
- b)forces on the particles
- c)relative distances between the particles
- d)position of the particles
Correct answer is option 'B'. Can you explain this answer?
The centre of mass of a system of particles does not depend on
a)
masses of the particles
b)
forces on the particles
c)
relative distances between the particles
d)
position of the particles
|
Neha Joshi answered |
The resultant of all forces, on any system of particles, is zero. Therefore, their centre of mass does not depend upon the forces acting on the particles.
If a shell at rest explodes then the centre of mass of the fragments
- a)remains at rest
- b)Moves along a parabolic path
- c)Moves along a straight line
- d)moves along an elliptical path
Correct answer is option 'A'. Can you explain this answer?
If a shell at rest explodes then the centre of mass of the fragments
a)
remains at rest
b)
Moves along a parabolic path
c)
Moves along a straight line
d)
moves along an elliptical path
|
Gaurav Kumar answered |
As a shell explosion is not governed by any external force, we get that no external force acts upon the shell, and hence it remains at rest.
If a man of mass M jumps to the ground from height h and his centre of mass moves a distance x in the time taken by him to hit the ground, the average force acting on him is (assuming his retardation to be constant during his impact with the ground)- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If a man of mass M jumps to the ground from height h and his centre of mass moves a distance x in the time taken by him to hit the ground, the average force acting on him is (assuming his retardation to be constant during his impact with the ground)
a)
b)
c)
d)
|
|
Neha Joshi answered |
As the center of mass moves through a distance x, the average force F acting on the man is calculated from :
Work done = Change in potential energy
⇒ Fx = Mgh
⇒ F =
Work done = Change in potential energy
⇒ Fx = Mgh
⇒ F =
When external forces acting on a body are zero, then its centre of mass- a)remains stationary
- b)moves with uniform velocity
- c)either remains stationary or moves with uniform velocity
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
When external forces acting on a body are zero, then its centre of mass
a)
remains stationary
b)
moves with uniform velocity
c)
either remains stationary or moves with uniform velocity
d)
none of these
|
|
Naina Sharma answered |
When force acting upon the body results zero, the resulting acceleration due to net force applied is also zero, and hence by the law of inertia the motion of the body either at rest or constant velocity wont change.
There are some passengers inside a stationary railway compartment. The centre of masses of the compartment itself(without the passengers) is C1, while the centre of mass of the compartment plus passengers’ system is C2. if the passengers moves about inside the compartment
- a)both C1 and C2 will move with respect to the ground
- b)neither C1 nor C2 will move with respect to the ground
- c)C1 will move but C2 will be stationary with respect to the ground
- d)C2 will move but C1 will be stationary with respect to the ground
Correct answer is option 'C'. Can you explain this answer?
There are some passengers inside a stationary railway compartment. The centre of masses of the compartment itself(without the passengers) is C1, while the centre of mass of the compartment plus passengers’ system is C2. if the passengers moves about inside the compartment
a)
both C1 and C2 will move with respect to the ground
b)
neither C1 nor C2 will move with respect to the ground
c)
C1 will move but C2 will be stationary with respect to the ground
d)
C2 will move but C1 will be stationary with respect to the ground
|
Lavanya Menon answered |
When net Fexternal=0, then the centre of mass of the system remains at rest.
Thus if the passenger move inside the compartment which donot require any external force, so the centre of mass of the "passenger + compartment" system must remain at rest and hence C2 will be fixed w.r.t ground.
Also due to the movement of the passenger, the position of centre of mass of the passengers only will change, thus C1 will have to move in such a way that C2 may remain fixed w.r.t ground.
A rigid body is one
- a)the sum of distances of all particles from the axis remains constant
- b)in which the distance between all pairs of particles remains fixed
- c)whose centre of mass follows a parabolic path
- d)that deforms and comes back to its original shape after getting deformed
Correct answer is option 'B'. Can you explain this answer?
A rigid body is one
a)
the sum of distances of all particles from the axis remains constant
b)
in which the distance between all pairs of particles remains fixed
c)
whose centre of mass follows a parabolic path
d)
that deforms and comes back to its original shape after getting deformed
|
|
Krishna Iyer answered |
A body is said to be a rigid body if the body remains in its original shape even under the influence of external force. We can also say that if distance between two points of the body does not change with time regardless of external forces exerted on it, then the body is said to be a rigid body.
A boy hits a baseball with a bat and imparts an impulse J to the ball. The boy hits the ball again with the same force, except that the ball and the bat are in contact for twice the amount of time as in the first hit. The new impulse equals.- a)half the original impulse
- b)the original impulse
- c)twice the original impulse
- d)four times the original impulse
Correct answer is option 'C'. Can you explain this answer?
A boy hits a baseball with a bat and imparts an impulse J to the ball. The boy hits the ball again with the same force, except that the ball and the bat are in contact for twice the amount of time as in the first hit. The new impulse equals.
a)
half the original impulse
b)
the original impulse
c)
twice the original impulse
d)
four times the original impulse
|
Hansa Sharma answered |
J = Ft
1) J = Ft1
2) J = F x 2t1
Hence impulse also gets doubled because impulse is dependent upon time
1) J = Ft1
2) J = F x 2t1
Hence impulse also gets doubled because impulse is dependent upon time
A uniform square plate ABCD has a mass of 10kg. If two point masses of 3 kg each are placed at the corners C and D as shown in the adjoining figure, then the centre of mass shifts to the point which is lie on- 
- a)OC
- b)OD
- c)OY
- d)OX
Correct answer is option 'C'. Can you explain this answer?
A uniform square plate ABCD has a mass of 10kg. If two point masses of 3 kg each are placed at the corners C and D as shown in the adjoining figure, then the centre of mass shifts to the point which is lie on-
a)
OC
b)
OD
c)
OY
d)
OX
|
Imk Pathsala answered |
∵ initial center of mass stays at the center of the square plate. When the mass (20 g) are placed at B and C, the shift along 'y' direction is Zero.
It moves towards the line OY
It moves towards the line OY
Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed v towards ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of v/ 3 in the same direction. What type of collision has occured ?- a)inelastic
- b)elastic
- c)completely inelastic
- d)Cannot be determined from the information given
Correct answer is option 'B'. Can you explain this answer?
Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed v towards ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of v/ 3 in the same direction. What type of collision has occured ?
a)
inelastic
b)
elastic
c)
completely inelastic
d)
Cannot be determined from the information given
|
|
Krishna Iyer answered |
First solve the equation of conservation of momentum which gives us that the relative velocity of approach is equal to the relative velocity of separation. Hence coefficient of restitution is 1. That means the collision is elastic.
mv2 + 2mv3 = 2mv + m0
- v2 = 4v3
- e = 4v3 − v3v – 0
- e = 1
Two particles having mass ratio n : 1 are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is :- a)(n _1)2 g
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Two particles having mass ratio n : 1 are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is :
a)
(n _1)2 g
b)
c)
d)
|
Learners Habitat answered |
Given 
Each mass will have the acceleration
However m1 which is heavier will have the will have acceleration a1 vertically down while the lighter mass m2 will have acceleration a2 vertically up → a2=−a1
The acceleration of the center of mass of the system,
Given that
Since
diving by m2 and simplifying 
Hence c is the correct answer
Each mass will have the acceleration
However m1 which is heavier will have the will have acceleration a1 vertically down while the lighter mass m2 will have acceleration a2 vertically up → a2=−a1
The acceleration of the center of mass of the system,
Given that
Since
diving by m2 and simplifying Hence c is the correct answer
A super-ball is to bounce elastically back and forth between two rigid walls at a distance d from each other. Neglecting gravity and assuming the velocity of super-ball to be v0 horizontally, the average force being exerted by the super-ball on each wall is :- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A super-ball is to bounce elastically back and forth between two rigid walls at a distance d from each other. Neglecting gravity and assuming the velocity of super-ball to be v0 horizontally, the average force being exerted by the super-ball on each wall is :
a)
b)
c)
d)
|
|
Naina Sharma answered |
Change in momentum when ball strikes one wall is:
Δp= −mv0−m(v0) = −2mv0
The time it takes to come back and hit the wall again is:
t=−2d/v0
Hence force applied is Δp/t=mv20/d
Δp= −mv0−m(v0) = −2mv0
The time it takes to come back and hit the wall again is:
t=−2d/v0
Hence force applied is Δp/t=mv20/d
Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is:
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is:
a)
b)
c)
d)
|
Sushil Kumar answered |
If, r1 and r2 are position vectors of the centres of the positional vector of their centre of mass is given by R=(m1r1+m2r2)/ (m1 + m2)
R= (m1r2+m2r2)/m1+m2)
The acceleration of the centre of mass is given by:
A=d2R/dt
= [m1 d2r/dt+m2 d2r/dt2 ]/(m1+m2)
But, d2r1/dt2 and d2r2/dt2 are the accelerations of masses m1 and m2
Have the same magnitude (m1-m2)/(m1+m2) g
If we take acceleration of [m1(m1-m2)/ (m1+m2)g – m2(m1-m2)/ (m1+m2)/]/ (m1+m2)
On simplifying that we get,
a=[(m1-m2)/ (m1+m2)]2 g
R= (m1r2+m2r2)/m1+m2)
The acceleration of the centre of mass is given by:
A=d2R/dt
= [m1 d2r/dt+m2 d2r/dt2 ]/(m1+m2)
But, d2r1/dt2 and d2r2/dt2 are the accelerations of masses m1 and m2
Have the same magnitude (m1-m2)/(m1+m2) g
If we take acceleration of [m1(m1-m2)/ (m1+m2)g – m2(m1-m2)/ (m1+m2)/]/ (m1+m2)
On simplifying that we get,
a=[(m1-m2)/ (m1+m2)]2 g
If three balls of same radius are placed touching each other on a horizontal surface such that they will form an equilateral triangle, when their centers are joined. What will be the position of the centre of mass of the system?- a)At the point of intersection of the medians.
- b)At the line joining the centers of any two balls.
- c)At the centre of one of the ball.
- d)At the horizontal surface.
Correct answer is option 'A'. Can you explain this answer?
If three balls of same radius are placed touching each other on a horizontal surface such that they will form an equilateral triangle, when their centers are joined. What will be the position of the centre of mass of the system?
a)
At the point of intersection of the medians.
b)
At the line joining the centers of any two balls.
c)
At the centre of one of the ball.
d)
At the horizontal surface.
|
|
Neha Joshi answered |
The COM will be at the intersection point of the medians as that is the most symmetric point of this symmetric system.
Two bodies of masses m1 and m2 are moving such that their momentums are equal and opposite. The velocity of centre of mass of system of two bodies is:- a)cannot be determined.
- b)equal to the velocity of particle m1
- c)zero
- d)equal to the velocity of particle m2
Correct answer is option 'C'. Can you explain this answer?
Two bodies of masses m1 and m2 are moving such that their momentums are equal and opposite. The velocity of centre of mass of system of two bodies is:
a)
cannot be determined.
b)
equal to the velocity of particle m1
c)
zero
d)
equal to the velocity of particle m2
|
|
Gaurav Kumar answered |
As momentum is p = mv and a vector quantity, if both are moving in opposite direction then the net momentum will be mv –mv. The minus sign shows that it is in opposite direction.
So p(net) is 0.
So p(net) is 0.
When a shell was following a parabolic path in the air, it explodes somewhere in its flight. The centre of mass of fragments will continue to move in- a)any direction
- b)horizontal direction
- c)same parabolic path
- d)vertical direction
Correct answer is option 'C'. Can you explain this answer?
When a shell was following a parabolic path in the air, it explodes somewhere in its flight. The centre of mass of fragments will continue to move in
a)
any direction
b)
horizontal direction
c)
same parabolic path
d)
vertical direction
|
|
Riya Banerjee answered |
The internal forces have no effect on the trajectory of the center of mass, and the forces due to explosion are the internal forces. So the center of mass will follow the same parabolic path even after the explosion.
Can you explain the answer of this question below:An isolated particle of mass m is moving in a horizontal plane (x,y) along the x axis at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragments is at y = +15 cm. The larger fragment at this instant is at
- A:
y = +5 cm
- B:
y = +20 cm
- C:
y = -20 cm
- D:
y = -5 cm
The answer is d.
An isolated particle of mass m is moving in a horizontal plane (x,y) along the x axis at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragments is at y = +15 cm. The larger fragment at this instant is at
y = +5 cm
y = +20 cm
y = -20 cm
y = -5 cm
|
Ambition Institute answered |
As the particle is exploded only due to its internal energy.
⇒ net external force during this process is 0 i.e. center mass will not change.
⇒ net external force during this process is 0 i.e. center mass will not change.
Let the particle while the explosion was above the origin of the coordinate system i.e. just before explosion xcm =0 and ycm =0
After the explosion, the Centre of mass will be at xcm =0 and ycm =0
Since smaller fragment has fallen on the y-axis.
Let positon of larger fragment be y.
Let positon of larger fragment be y.
m * ycm = (m/4 * 15) + (3m/4 * y)
⇒ (m/4 * 15) + (3m/4 * y) = 0
⇒ y = - 5 cm
⇒ (m/4 * 15) + (3m/4 * y) = 0
⇒ y = - 5 cm
A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal to- a)
- b)
- c)
- d)3R
Correct answer is option 'C'. Can you explain this answer?
A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal to
a)
b)
c)
d)
3R
|
|
Nandini Iyer answered |
At the time when the particle broke the momentum is conserved as no external force acted
As the vertical component was zero and let say horizontal component was v
Thus we get,
3mv = m.0 + 2m.u
Thus we get u = 3/2 . v
But as there was no change in vertical component of velocity, time of flight does not change. Thus the distance at which the particle would land would be t x v + t x u
Where 2t is time of flight and 2t x v = R
Thus x = R/2 + 3R/4
= 5R/4
As the vertical component was zero and let say horizontal component was v
Thus we get,
3mv = m.0 + 2m.u
Thus we get u = 3/2 . v
But as there was no change in vertical component of velocity, time of flight does not change. Thus the distance at which the particle would land would be t x v + t x u
Where 2t is time of flight and 2t x v = R
Thus x = R/2 + 3R/4
= 5R/4
A ball is dropped from a height h. As is bounces off the floor, its speed is 80 percent of what it was just before it hit the floor. The ball will then rise to a height of most nearly- a)0.80 h
- b)0.75 h
- c)0.64 h
- d)0.50 h
Correct answer is option 'C'. Can you explain this answer?
A ball is dropped from a height h. As is bounces off the floor, its speed is 80 percent of what it was just before it hit the floor. The ball will then rise to a height of most nearly
a)
0.80 h
b)
0.75 h
c)
0.64 h
d)
0.50 h
|
|
Lavanya Menon answered |
Mgh = 1/2mu2
If v = 0.8u
½ × 0.8u2 × m = mgh'
So,h' = 0.82h
h' = 0.64h
If v = 0.8u
½ × 0.8u2 × m = mgh'
So,h' = 0.82h
h' = 0.64h
A bomb initially at rest explodes by it self into three equal mass fragments. The velocities of two fragments are ( 3 
+ 2 
) m/s and (_
_ 4
) m/s. The velocity of the third fragment is (in m/s)-- a)2
+ 2
- b)2
_ 2
- c)_ 2
+ 2
- d)_2
_ 2
Correct answer is option 'C'. Can you explain this answer?
A bomb initially at rest explodes by it self into three equal mass fragments. The velocities of two fragments are ( 3 + 2 ) m/s and (_ _ 4) m/s. The velocity of the third fragment is (in m/s)-
+ 2 ) m/s and (_ _ 4) m/s. The velocity of the third fragment is (in m/s)-a)
2 + 2
+ 2b)
2 _ 2
_ 2c)
_ 2 + 2
+ 2d)
_2 _ 2
_ 2|
|
EduRev JEE answered |
Applying law of conservation of momentum,
Mv= m1v1 + m2v2 +m3v3
As initially M is at rest ,v= O
Also m1=m2 =m3 = m (equal masses)
We have m(v1 +v2+ v3) = O
v1+ v2 = -v3
(3i+2j)+(-i-4j) = -v3
3i+2j-i-4j =2i-2j =-v3
v3 = -2i+2j
Mv= m1v1 + m2v2 +m3v3
As initially M is at rest ,v= O
Also m1=m2 =m3 = m (equal masses)
We have m(v1 +v2+ v3) = O
v1+ v2 = -v3
(3i+2j)+(-i-4j) = -v3
3i+2j-i-4j =2i-2j =-v3
v3 = -2i+2j
Which of the following options are correct,
where i, j and k are unit vectors along the x, y and z axis?- a)i.j = 1 ; j x i =0
- b)i.j = 0 ; j x i = -k
- c)i.j = 1 ; j x i = -k
- d)i.j = 1 ; j x i = k
Correct answer is option 'B'. Can you explain this answer?
Which of the following options are correct,
where i, j and k are unit vectors along the x, y and z axis?
where i, j and k are unit vectors along the x, y and z axis?
a)
i.j = 1 ; j x i =0
b)
i.j = 0 ; j x i = -k
c)
i.j = 1 ; j x i = -k
d)
i.j = 1 ; j x i = k
|
|
Hansa Sharma answered |
Dot product of two different unit vectors is 0 and dot product of two same unit vectors is 1. Cross product of two different unit vectors taken according to right hand thumb rule is the other vector. Cross product of two same unit vectors is 0.
If the net force acting on the system of particles is zero, then which of the following may vary
- a)moment of inertia
- b)Kinetic energy of the system
- c)velocity of centre of mass
- d)position of centre of mass
Correct answer is option 'D'. Can you explain this answer?
If the net force acting on the system of particles is zero, then which of the following may vary
a)
moment of inertia
b)
Kinetic energy of the system
c)
velocity of centre of mass
d)
position of centre of mass
|
|
Suresh Kumar answered |
Correct answer is B . because when a=0 then v=constant hence K.E. will be constant and position of com will vary
A metal ball hits a wall and does not rebound whereas a rubber ball of the same mass on hitting the wall the same velocity rebounds back. It can be concluded that -- a)metal ball sufferes greater change in momentum
- b)rubber ball suffers greater change in momentum
- c)the initial momentum of metal ball is greater than the initial momentum of rubber ball
- d)both suffer same change in momentum
Correct answer is option 'B'. Can you explain this answer?
A metal ball hits a wall and does not rebound whereas a rubber ball of the same mass on hitting the wall the same velocity rebounds back. It can be concluded that -
a)
metal ball sufferes greater change in momentum
b)
rubber ball suffers greater change in momentum
c)
the initial momentum of metal ball is greater than the initial momentum of rubber ball
d)
both suffer same change in momentum
|
|
Geetika Shah answered |
According to question metal ball doesn't rebound and rubber ball of same mass rebounds, so, rubber ball has change in momentum at least twice as that of metal ball.
mv - (-mv) = 2mv
mv - (-mv) = 2mv
A child sits stationary at one end of long trolley moving uniformly with speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in the forward direction with speed u. The centre of mass of the system (child + trolley) will move with speed- a)v
- b)zero
- c)u + v
- d)v/u
Correct answer is option 'A'. Can you explain this answer?
A child sits stationary at one end of long trolley moving uniformly with speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in the forward direction with speed u. The centre of mass of the system (child + trolley) will move with speed
a)
v
b)
zero
c)
u + v
d)
v/u
|
Preeti Iyer answered |
The child is running arbitrarily on a trolley moving with velocity v. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy–trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.
An object of mass 5 kg and speed 10 ms-1 explodes into two pieces of equal mass. One piece comes to rest. The kinetic energy added to the system during the explosion is-- a)Zero
- b)50 J
- c)250 J
- d)500 J
Correct answer is option 'C'. Can you explain this answer?
An object of mass 5 kg and speed 10 ms-1 explodes into two pieces of equal mass. One piece comes to rest. The kinetic energy added to the system during the explosion is-
a)
Zero
b)
50 J
c)
250 J
d)
500 J
|
|
Suresh Iyer answered |
ATQ: Given value mass of object = 5 kg and v = 10 m/s
K.E = 1/2 mv2
1/2 × 5 × 100 = 250 J
K.E = 1/2 mv2
1/2 × 5 × 100 = 250 J
A stone of mass m1 moving with a uniform speed v suddenly explodes on its own into two fragments. If the fragment of mass m2 is at rest, the speed of the other fragment is-- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A stone of mass m1 moving with a uniform speed v suddenly explodes on its own into two fragments. If the fragment of mass m2 is at rest, the speed of the other fragment is-
a)
b)
c)
d)
|
|
Geetika Shah answered |
As the stone of mass m1 moving with a constant speed, v,
its momentum is m1v.
When it explodes into 2 fragments, we have m2 the fragment, which remains at rest and hence, its momentum is 0.
The mass of other fragment is m1−m2, if its velocity is V, then its momentum is V(m1−m2).
As momentum is conserved, we should have momentum before the split and after the split as equal and therefore
m1v=0+V(m1−m2)
or, V(m1−m2) =m1v
or, V=m1v/m1−m2
its momentum is m1v.
When it explodes into 2 fragments, we have m2 the fragment, which remains at rest and hence, its momentum is 0.
The mass of other fragment is m1−m2, if its velocity is V, then its momentum is V(m1−m2).
As momentum is conserved, we should have momentum before the split and after the split as equal and therefore
m1v=0+V(m1−m2)
or, V(m1−m2) =m1v
or, V=m1v/m1−m2
In an inelastic collision-- a)momentum is conserved but kinetic energy is not consereved
- b)momentum is not conserved but kinetic energy is conserved
- c)neighter momentum nor kinetic energy is conserved
- d)both the momentum and kinetic energy are conserved
Correct answer is option 'A'. Can you explain this answer?
In an inelastic collision-
a)
momentum is conserved but kinetic energy is not consereved
b)
momentum is not conserved but kinetic energy is conserved
c)
neighter momentum nor kinetic energy is conserved
d)
both the momentum and kinetic energy are conserved
|
|
Neha Joshi answered |
In an inelastic collision momentum is conserved but kinetic energy is not conserved, by the virtue of its definition.
A force exerts an impulse I on a particle changing its speed from u to 2u. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is- a)
- b)
- c)Iu
- d)2Iu
Correct answer is option 'B'. Can you explain this answer?
A force exerts an impulse I on a particle changing its speed from u to 2u. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is
a)
b)
c)
Iu
d)
2Iu
|
|
Hansa Sharma answered |
Impulse, I=change of momentum=2mu−(−mu)=3mu
Work done, W= change of kinetic energy=1/2m(2u)2−1/2mu2
=3/2mu2
=1/2Iu
Work done, W= change of kinetic energy=1/2m(2u)2−1/2mu2
=3/2mu2
=1/2Iu
When two bodies collide elastically, then- a)kinetic energy of the system alone is conserved
- b)only momentum is conserved
- c)both energy and momentum are conserved
- d)neighter energy nor momentum is conserved
Correct answer is option 'C'. Can you explain this answer?
When two bodies collide elastically, then
a)
kinetic energy of the system alone is conserved
b)
only momentum is conserved
c)
both energy and momentum are conserved
d)
neighter energy nor momentum is conserved
|
|
Suresh Reddy answered |
In collisions in absence of any external force the momentum is conserved. As the collision is elastic so there will not be any deformation in the shapes of the bodies so there will not be any loss in energy so the energy will remain constant.
A ball is dropped from height 5m. The time after which ball stops rebounding if coefficient of restitution between ball and ground e = 1/2, is- a)1 sec
- b)2 sec
- c)3 sec
- d)infinite
Correct answer is option 'C'. Can you explain this answer?
A ball is dropped from height 5m. The time after which ball stops rebounding if coefficient of restitution between ball and ground e = 1/2, is
a)
1 sec
b)
2 sec
c)
3 sec
d)
infinite
|
|
Suresh Reddy answered |
The speed of ball just before the first rebound = 
Speed of ball just after first rebound = 10 x ½
= 5 m/s
Time taken by the ball before first rebound is 1 sec
Similarly time taken between first and second rebound is 2( 5/10) = ½ sec
Speed of ball after second rebound = 5 x ½
= 2.5 m/s
Hence the time between second and third rebound is ¼ second
Similarly if we find time between more succeeding rebounds we get an infinite GP of common ratio ½ .
Hence the sum is 2 sec.
Speed of ball just after first rebound = 10 x ½
= 5 m/s
Time taken by the ball before first rebound is 1 sec
Similarly time taken between first and second rebound is 2( 5/10) = ½ sec
Speed of ball after second rebound = 5 x ½
= 2.5 m/s
Hence the time between second and third rebound is ¼ second
Similarly if we find time between more succeeding rebounds we get an infinite GP of common ratio ½ .
Hence the sum is 2 sec.
A ball hits the floor and rebounds after an inelastic collision . In this case-- a)the momentum of the ball just after the collision is the same as that just before the collision
- b)the mechanical energy of the ball remains the same in the collision
- c)the total momentum of the ball and the earth is conserved.
- d)the total energy of the ball and the earth is conserved
Correct answer is option 'C'. Can you explain this answer?
A ball hits the floor and rebounds after an inelastic collision . In this case-
a)
the momentum of the ball just after the collision is the same as that just before the collision
b)
the mechanical energy of the ball remains the same in the collision
c)
the total momentum of the ball and the earth is conserved.
d)
the total energy of the ball and the earth is conserved
|
|
Neha Joshi answered |
When the collision is inelastic the ball doesn't rebond at all but it will stick to the ground. So the momentum of the ball just before and after collision is not conserved and so is it's energy. While considering both the ball and ground as our system the total energy is not conserved since ball losses it's energy but the ground doesn't gain any energy. However momentum will be now conserved because there is no external force acting on our system (gravitational force becomes internal).
An object comprises of a uniform ring of radius R and its uniform chord AB (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object.
- a) (R/3, R/3)
- b) (R/3, R/2)
- c)(R/4, R/4)
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
An object comprises of a uniform ring of radius R and its uniform chord AB (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object.
a)
(R/3, R/3)
b)
(R/3, R/2)
c)
(R/4, R/4)
d)
None of these
|
|
Raghav Bansal answered |
Since the points must have symmetric coordinates so B is the answer.
A ball strikes a smooth horizontal ground at an angle of 45° with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision. (Assume e <=1).- a)45°
- b)30°
- c)53°
- d)60°
Correct answer is option 'B'. Can you explain this answer?
A ball strikes a smooth horizontal ground at an angle of 45° with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision. (Assume e <=1).
a)
45°
b)
30°
c)
53°
d)
60°
|
|
Rahul Kumar answered |
Degrees with the horizontal. The initial velocity of the ball is 20 m/s. Determine the time it takes for the ball to hit the ground and the horizontal distance it travels before hitting the ground.
To solve this problem, we can break the initial velocity of the ball into its horizontal and vertical components. Since the angle of projection is 45 degrees, the horizontal and vertical components will be equal.
Horizontal component of velocity (Vx) = initial velocity (V) * cos(angle)
Vx = 20 m/s * cos(45 degrees)
Vx = 20 m/s * 0.7071
Vx ≈ 14.14 m/s
Vertical component of velocity (Vy) = initial velocity (V) * sin(angle)
Vy = 20 m/s * sin(45 degrees)
Vy = 20 m/s * 0.7071
Vy ≈ 14.14 m/s
Now, we can analyze the vertical motion of the ball. The ball is thrown upwards, reaches its highest point, and then falls back down. At the highest point, the vertical velocity becomes zero.
Using the equation of motion for vertical motion:
Vy = initial vertical velocity (Uy) + acceleration due to gravity (g) * time (t)
0 = 14.14 m/s - 9.8 m/s^2 * t
Solving for t:
t = 14.14 m/s / 9.8 m/s^2
t ≈ 1.44 seconds
Now, we can determine the horizontal distance traveled by the ball using the horizontal component of velocity (Vx) and the time of flight (t).
Horizontal distance (D) = horizontal velocity (Vx) * time (t)
D = 14.14 m/s * 1.44 seconds
D ≈ 20.38 meters
Therefore, it takes approximately 1.44 seconds for the ball to hit the ground, and it travels approximately 20.38 meters horizontally before hitting the ground.
To solve this problem, we can break the initial velocity of the ball into its horizontal and vertical components. Since the angle of projection is 45 degrees, the horizontal and vertical components will be equal.
Horizontal component of velocity (Vx) = initial velocity (V) * cos(angle)
Vx = 20 m/s * cos(45 degrees)
Vx = 20 m/s * 0.7071
Vx ≈ 14.14 m/s
Vertical component of velocity (Vy) = initial velocity (V) * sin(angle)
Vy = 20 m/s * sin(45 degrees)
Vy = 20 m/s * 0.7071
Vy ≈ 14.14 m/s
Now, we can analyze the vertical motion of the ball. The ball is thrown upwards, reaches its highest point, and then falls back down. At the highest point, the vertical velocity becomes zero.
Using the equation of motion for vertical motion:
Vy = initial vertical velocity (Uy) + acceleration due to gravity (g) * time (t)
0 = 14.14 m/s - 9.8 m/s^2 * t
Solving for t:
t = 14.14 m/s / 9.8 m/s^2
t ≈ 1.44 seconds
Now, we can determine the horizontal distance traveled by the ball using the horizontal component of velocity (Vx) and the time of flight (t).
Horizontal distance (D) = horizontal velocity (Vx) * time (t)
D = 14.14 m/s * 1.44 seconds
D ≈ 20.38 meters
Therefore, it takes approximately 1.44 seconds for the ball to hit the ground, and it travels approximately 20.38 meters horizontally before hitting the ground.
Three identical spheres each of radius R are placed such that their centres lie on a straight line. What is the location of their centre of mass from the centre of the first sphere?
- a)R
- b)2R
- c)3R
- d)4R
Correct answer is option 'B'. Can you explain this answer?
Three identical spheres each of radius R are placed such that their centres lie on a straight line. What is the location of their centre of mass from the centre of the first sphere?
a)
R
b)
2R
c)
3R
d)
4R
|
|
Geetika Shah answered |
- Distance between first and last sphere = R + 2R + R = 4R
- Since the spheres are identical and lie in a straight line, the centre of mass will lie exactly in the middle.
- Hence the centre of mass lies at a distance of 2R from the centre of the first sphere.
The sum of moments of masses of all the particles in a system about the centre of mass is always:- a)maximum
- b)infinite
- c)zero
- d)minimum
Correct answer is option 'C'. Can you explain this answer?
The sum of moments of masses of all the particles in a system about the centre of mass is always:
a)
maximum
b)
infinite
c)
zero
d)
minimum
|
|
Neha Sharma answered |
In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero, or the point where if a force is applied it moves in the direction of the force without rotating. The distribution of mass is balanced around the center of mass and the average of the weighted position coordinates of the distributed mass defines its coordinates. Calculations in mechanics are often simplified when formulated with respect to the center of mass.
There are two particles of masses m and 2m placed at a distance‘d’ apart on a smooth horizontal surface. Where will the collision occur (with respect to original positions), if they are allowed to move towards each other because of their mutual attraction?- a)2d/3
- b)d/2
- c)2d
- d)d
Correct answer is option 'A'. Can you explain this answer?
There are two particles of masses m and 2m placed at a distance‘d’ apart on a smooth horizontal surface. Where will the collision occur (with respect to original positions), if they are allowed to move towards each other because of their mutual attraction?
a)
2d/3
b)
d/2
c)
2d
d)
d
|
|
Pooja Shah answered |
Their collision will occur at their center of mass.
If we consider body of mass 'm' at origin & '2m' at distance "d" from 1st body on X-axis then, center of mass,
X = m1(x1)+m2(x2)/m1+m2
X = m(0)+2m(d)/m+2m
X = 2md/3m
X = 2d/3
Hence A is the correct answer.
If we consider body of mass 'm' at origin & '2m' at distance "d" from 1st body on X-axis then, center of mass,
X = m1(x1)+m2(x2)/m1+m2
X = m(0)+2m(d)/m+2m
X = 2md/3m
X = 2d/3
Hence A is the correct answer.
When a fire cracker initially at rest, explodes into a number of fragments, the centre of mass- a)remains stationary
- b)moves along a parabolic path
- c)moves along vertical direction
- d)moves along horizontal direction
Correct answer is option 'A'. Can you explain this answer?
When a fire cracker initially at rest, explodes into a number of fragments, the centre of mass
a)
remains stationary
b)
moves along a parabolic path
c)
moves along vertical direction
d)
moves along horizontal direction
|
Om Desai answered |
When a fire cracker initially at rest, explodes into a number of fragments, the centre of mass remains stationary. This is analogous to the explosion of a ball while it is flying in a trajectory.
A truck moving on horizontal road east with velocity 20ms-1 collides elastically with a light ball moving with velocity 25 ms-1 along west. The velocity of the ball just after collision- a)65 ms-1 towards east
- b)25 ms-1 towards west
- c)65 ms-1 towards west
- d)20 ms-1 towards east
Correct answer is option 'A'. Can you explain this answer?
A truck moving on horizontal road east with velocity 20ms-1 collides elastically with a light ball moving with velocity 25 ms-1 along west. The velocity of the ball just after collision
a)
65 ms-1 towards east
b)
25 ms-1 towards west
c)
65 ms-1 towards west
d)
20 ms-1 towards east
|
|
Om Desai answered |
Since the system is under no external force momentum is conserved.
Let the mass of truck be M and ball be m, ATQ M>>m
Now, 20M – 25m = mv2 – Mv1 --- (1) where v1 and v2 are the final velocities of the truck and ball respectively
Also, e = ((v2) - (-v1))/((20) - (-25)) = 1
So, v1 + v2 = 45 --- (2)
From equation (1)
20 – 25m/M = m/M v2 – v1
-20 = v1
So, from (2)
Finally, v2 = 65m/s along east
Let the mass of truck be M and ball be m, ATQ M>>m
Now, 20M – 25m = mv2 – Mv1 --- (1) where v1 and v2 are the final velocities of the truck and ball respectively
Also, e = ((v2) - (-v1))/((20) - (-25)) = 1
So, v1 + v2 = 45 --- (2)
From equation (1)
20 – 25m/M = m/M v2 – v1
-20 = v1
So, from (2)
Finally, v2 = 65m/s along east
Every point in a rotating rigid body has the same __________at any instant of time.
- a)linear velocity
- b)angular velocity
- c)displacement
- d)linear momentum
Correct answer is option 'B'. Can you explain this answer?
Every point in a rotating rigid body has the same __________at any instant of time.
a)
linear velocity
b)
angular velocity
c)
displacement
d)
linear momentum
|
|
Preeti Iyer answered |
Every point in any rigid rotating object is rotating at the same angular velocity. However, there are a few cases where people have used the term "angular velocity" when they really meant tangential velocity, so you do have to be careful.
The general motion of a rigid body consists of
- a)translational motion only
- b)rotational motion only
- c)both, translational and rotational motion
- d)does not include translational and rotational motions
Correct answer is option 'C'. Can you explain this answer?
The general motion of a rigid body consists of
a)
translational motion only
b)
rotational motion only
c)
both, translational and rotational motion
d)
does not include translational and rotational motions
|
|
Arun Khanna answered |
The general motion of a rigid body consists of both, translational and rotational motion. This is much obvious as it is seen all around us.
A ball of 0.1kg strikes a wall at right angle with a speed of 6 m/s and rebounds along its original path at 4 m/s. The change in momentum in Newton- sec is-- a)103
- b)102
- c)10
- d)1
Correct answer is option 'D'. Can you explain this answer?
A ball of 0.1kg strikes a wall at right angle with a speed of 6 m/s and rebounds along its original path at 4 m/s. The change in momentum in Newton- sec is-
a)
103
b)
102
c)
10
d)
1
|
Moumita Chakraborty answered |
Mass of the ball = 0.1 Kg
ui = u = 6 m/s
uf = v = -4 m/s
∴ ΔP = | mΔv |
= | m (v-u) |
= | 0.1 ( -4 - 6 ) |
= | 0.1 ( -10 ) |
= 1
Two balls are thrown simultaneously in air. The acceleration of the centre of mass of the two balls, while in air- a)depend on the speeds of the two balls
- b)depend on the masses of the two balls
- c)depend on the direction of the motion of the balls
- d)is equal to g
Correct answer is option 'D'. Can you explain this answer?
Two balls are thrown simultaneously in air. The acceleration of the centre of mass of the two balls, while in air
a)
depend on the speeds of the two balls
b)
depend on the masses of the two balls
c)
depend on the direction of the motion of the balls
d)
is equal to g
|
Mira Sharma answered |
The only force acting on the balls is force of gravity, so the acceleration of centre of mass is equal to g.
A proton and an electron, initially at rest, are allowed to move under their mutal attractive force. Their centre of mass will:- a)move towards the proton
- b)move towards the electron
- c)move in an unpredictable manner
- d)remain stationary
Correct answer is option 'D'. Can you explain this answer?
A proton and an electron, initially at rest, are allowed to move under their mutal attractive force. Their centre of mass will:
a)
move towards the proton
b)
move towards the electron
c)
move in an unpredictable manner
d)
remain stationary
|
|
Arya Dasgupta answered |
Explanation:
When a proton and an electron are allowed to move under their mutual attractive force, their centre of mass remains stationary. This can be explained by the following factors:
1. Conservation of momentum: According to Newton's third law, the forces acting on the proton and electron are equal and opposite. Therefore, the momentum gained by the proton is equal and opposite to the momentum gained by the electron. As a result, the total momentum of the system remains zero, which means the centre of mass remains stationary.
2. Mass ratio: The mass of the proton is much larger than the mass of the electron. Therefore, the proton moves much less than the electron towards the centre of mass. This means that the centre of mass is closer to the proton, but still remains stationary.
3. Distance between the particles: The force of attraction between the proton and electron is inversely proportional to the square of the distance between them. As the particles move towards each other, the force of attraction increases, which causes an acceleration in both particles. However, since the force also decreases as the particles get closer, the acceleration decreases. This means that the particles move towards each other at an ever-decreasing speed, and eventually come to a stop at a fixed distance from each other. At this point, the centre of mass remains stationary.
Therefore, the correct answer is option 'D': the centre of mass remains stationary.
When a proton and an electron are allowed to move under their mutual attractive force, their centre of mass remains stationary. This can be explained by the following factors:
1. Conservation of momentum: According to Newton's third law, the forces acting on the proton and electron are equal and opposite. Therefore, the momentum gained by the proton is equal and opposite to the momentum gained by the electron. As a result, the total momentum of the system remains zero, which means the centre of mass remains stationary.
2. Mass ratio: The mass of the proton is much larger than the mass of the electron. Therefore, the proton moves much less than the electron towards the centre of mass. This means that the centre of mass is closer to the proton, but still remains stationary.
3. Distance between the particles: The force of attraction between the proton and electron is inversely proportional to the square of the distance between them. As the particles move towards each other, the force of attraction increases, which causes an acceleration in both particles. However, since the force also decreases as the particles get closer, the acceleration decreases. This means that the particles move towards each other at an ever-decreasing speed, and eventually come to a stop at a fixed distance from each other. At this point, the centre of mass remains stationary.
Therefore, the correct answer is option 'D': the centre of mass remains stationary.
Chapter doubts & questions for Centre of Mass - Physics for EmSAT Achieve 2025 is part of EmSAT Achieve exam preparation. The chapters have been prepared according to the EmSAT Achieve exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for EmSAT Achieve 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Centre of Mass - Physics for EmSAT Achieve in English & Hindi are available as part of EmSAT Achieve exam.
Download more important topics, notes, lectures and mock test series for EmSAT Achieve Exam by signing up for free.
Physics for EmSAT Achieve
208 videos|329 docs|212 tests
|
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup