Test: Centre of Mass - 3 - JEE MCQ

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

Mass of the ball = 0.1 Kg 

u_i  = u = 6 m/s

u_f = v = -4 m/s

∴ ΔP = | mΔv |

         = | m (v-u) |

         = | 0.1 ( -4 - 6 ) |

         = | 0.1 ( -10 ) |

          = 1

Obviously the conservation of energy is valid and the direct consequence of it is that KE is converted to compressed PE

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

ATQ: Given value mass of object = 5 kg and v = 10 m/s
K.E = 1/2 mv²
1/2 × 5 × 100 = 250 J

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.

Mgh = 1/2mu²
If v = 0.8u
½ × 0.8u2 × m = mgh^'
So,h' = 0.8²h
h' = 0.64h

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

Assuming external forces = 0
1 x 21 – 2 x 4 = 1 x 1 + 2 x v
So, v = 6m/s along the direction of A
Now, e = ((6) - (1))/((21) - (-4))
Hence e = 0.2

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.

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).

In an inelastic collision momentum is conserved but kinetic energy is not conserved, by the virtue of its definition.

The x component will remain same while the y component will become ‘e’ times.

Now to find the new angle

tanα = 1/e
Since, e<=1 tanα >= 1 so α >= initial angle

J = Ft
1) J = Ft1
2) J = F x 2t1
Hence impulse also gets doubled because impulse is dependent upon time

Conservative of linear moment is when there is no external force on body and this way the case of action and reaction pair cancelling each other by Newton third law of motion.

Let Xm be the displacement of man and Xp of plank
So, Xcom = (MXm + M/3 (-Xp))/(4M/3)
Since, Xcom = 0
So, 3Xm = Xp
Also, Xm + Xp = L

∵ 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
 

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

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

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

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

From law of conservation of momentum, we have momentum before collision = momentum after collision
The daughter nucleus has mass

The daughter nucleus has mass

The negative sign represents recoil.
Hence A is the correct answer.

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

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