Previous year questions (2016-20) - Motion Notes | EduRev

Physics For JEE

JEE : Previous year questions (2016-20) - Motion Notes | EduRev

The document Previous year questions (2016-20) - Motion Notes | EduRev is a part of the JEE Course Physics For JEE.
All you need of JEE at this link: JEE

Q.1.  A particle is moving along the x-axis, with its coordinate as a function of time given by x(t) =10+ 8t - 3t2. Another particle is moving along the y-axis with its coordinate as a function of time given by y(t) = 5 - 8t3. At t = 1s, the speed of the second particle as measured in the frame of the first particle is given as √v. Then v (in m/s) is ______.    [2020]
Ans.
580
Solution: Given that
 x(t) =10+ 8t - 3t2, y(t) = 5 - 8t3 
So,
Previous year questions (2016-20) - Motion Notes | EduRev
At t = 1 s, we have
Previous year questions (2016-20) - Motion Notes | EduRev
Relative velocity of second particle with respect to first particle is
Previous year questions (2016-20) - Motion Notes | EduRev

Q.2. A particle moves such that it's position vector is Previous year questions (2016-20) - Motion Notes | EduRev where ω is a constant and t is time. Then which of the following statements is true for the velocity Previous year questions (2016-20) - Motion Notes | EduRev and acceleration  Previous year questions (2016-20) - Motion Notes | EduRev of the particle.
(1) Previous year questions (2016-20) - Motion Notes | EduRev is perpendicular to Previous year questions (2016-20) - Motion Notes | EduRevis directed away from the origin.
(2) Previous year questions (2016-20) - Motion Notes | EduRev both are perpendicular to Previous year questions (2016-20) - Motion Notes | EduRev
(3) Previous year questions (2016-20) - Motion Notes | EduRev both are parallel to Previous year questions (2016-20) - Motion Notes | EduRev
(4) Previous year questions (2016-20) - Motion Notes | EduRev is perpendicular to  Previous year questions (2016-20) - Motion Notes | EduRev is directed towards the origin.
Ans:
(4)
Solution: Given that
Previous year questions (2016-20) - Motion Notes | EduRev
So,
Previous year questions (2016-20) - Motion Notes | EduRev
and
Previous year questions (2016-20) - Motion Notes | EduRev
Since, Previous year questions (2016-20) - Motion Notes | EduRev is perpendicular to Previous year questions (2016-20) - Motion Notes | EduRev
And Previous year questions (2016-20) - Motion Notes | EduRev had opposite direction to  Previous year questions (2016-20) - Motion Notes | EduRev or antiparallel to  Previous year questions (2016-20) - Motion Notes | EduRev that is, directed towards the origin.

Q.3. A ball is dropped from the top of a 100 m high tower on a planet. In the last (1/2)s before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in m/s2) near the surface on that planet is _______.
Ans.
8
Solution: Given that
u = 0 m/s, H = 100 m, d = 19 m, t = 1/2 s
Let time taken to tavel (H – h) distance be t.
So,
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev    ..(1)
Now, time taken to hit the ground is
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev    ..(2)
From Eqs. (1) and (2), we get
Previous year questions (2016-20) - Motion Notes | EduRev
⇒ a = 8 m/s2


Q.4. A particle starts from the origin at t = 0 with an initial velocity of Previous year questions (2016-20) - Motion Notes | EduRev and moves in the x-y plane with a constant acceleration Previous year questions (2016-20) - Motion Notes | EduRev The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D m. The value of D is
(1) 32
(2) 50
(3) 60
(4) 40
Ans.
(3)
Solution: Given that
Previous year questions (2016-20) - Motion Notes | EduRev
From the equation of motion in two dimensions, we have
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
⇒  t2 = 16
⇒ t = 4 s
So,
Previous year questions (2016-20) - Motion Notes | EduRev

Q.5. The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x−n, where n is an integer, the value of n is ______.
Ans.
3
Solution: x2 = at2 + 2bt + c
Previous year questions (2016-20) - Motion Notes | EduRev    .. (1)
Velocity is given by
Previous year questions (2016-20) - Motion Notes | EduRev
⇒ vx = at + b    ..(2)
Acceleration is given by
Previous year questions (2016-20) - Motion Notes | EduRev
Differentiate Eq. (2) with respect to t, we get
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
So,
Previous year questions (2016-20) - Motion Notes | EduRev
Therefore, n = 3

Q 6. In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed v more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2, respectively. Then v is equal to    [JEE Main 2019]
(1) Previous year questions (2016-20) - Motion Notes | EduRev
(2) Previous year questions (2016-20) - Motion Notes | EduRev
(3) Previous year questions (2016-20) - Motion Notes | EduRev
(4) 
Previous year questions (2016-20) - Motion Notes | EduRev
Ans: (3)
Solution: Let time taken by car A is t'
Previous year questions (2016-20) - Motion Notes | EduRev
vB = a2(t' +t)
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Put this value in Eq. (1), we get
Previous year questions (2016-20) - Motion Notes | EduRev

Q 7. A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of the. The ratio of times taken by the passenger train to completely cross the freight train when (i) they are moving in same direction and (ii) in the opposite directions is  [JEE Main 2019]
(1) 11/5
(2) 5/2
(3) 3/2
(4) 25/11
Ans:
(1)
Solution: 
Let passenger train and freight train be A and B and vA is the velocity of passenger train and vB is the velocity of freight train B.

Relative velocity of A with respect to B is (same direction)
Previous year questions (2016-20) - Motion Notes | EduRev
Time taken by the train A to cross the train B
Previous year questions (2016-20) - Motion Notes | EduRev
Train moving in opposite direction
Previous year questions (2016-20) - Motion Notes | EduRev
Therefore, required ratio is
Previous year questions (2016-20) - Motion Notes | EduRev

Q 8. A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5 s?    (JEE Main 2019)
Previous year questions (2016-20) - Motion Notes | EduRev
(1) 10 m
(2) 6 m
(3) 3 m
(4) 9 m
Ans:
(4)
As we know that, area under v-t curve gives displacement.
therefore,
Previous year questions (2016-20) - Motion Notes | EduRev

Q 9. A particle is moving with a velocity Previous year questions (2016-20) - Motion Notes | EduRev where K is a constant. The general equation for its path is    (JEE Main 2019)
(1) y = x2 + constant
(2) y2 = x + constant
(3) y2 = x2 + constant
(4) xy = constant
Ans:
(3)
Solution: Equation for velocity of the particle is
Previous year questions (2016-20) - Motion Notes | EduRev
Thus,
Previous year questions (2016-20) - Motion Notes | EduRev
Now,
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Integrating both the sides, we get
y2 = x2 + constant

Q 10.  The position coordinates of a particle moving in a 3D coordinate system is given by
x = a cos ωt
y = a sin ωt
and
z = aωt
The speed of the particle is    (JEE Main 2019)
(1) Previous year questions (2016-20) - Motion Notes | EduRev
(2) aω
(3) Previous year questions (2016-20) - Motion Notes | EduRev
(4) 2aω
Ans:
(1)
Solution:
Velocity coordinates of particle is obtained by differentiating the position coordinate with respect to time, Thus:
Previous year questions (2016-20) - Motion Notes | EduRev 
Hence, velocity of particle is
Previous year questions (2016-20) - Motion Notes | EduRev

Q 11. In the cube of side a, as shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be    (JEE Main 2019)
Previous year questions (2016-20) - Motion Notes | EduRev
(1) Previous year questions (2016-20) - Motion Notes | EduRev
(2) Previous year questions (2016-20) - Motion Notes | EduRev
(3) Previous year questions (2016-20) - Motion Notes | EduRev
(4) Previous year questions (2016-20) - Motion Notes | EduRev
Ans:
(3)
Solution:
Side of a cube is a
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev

Q 12. Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is     (JEE Main 2019)
(1) 1 : 16
(2) 1 : 2
(3) 1 : 4
(4) 1 : 8
Ans: 
(1)
Solution:
Area covered = πR2 

A ∝ R

Thus, range
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Therefore, required ratio is
Previous year questions (2016-20) - Motion Notes | EduRev

Q 13. Two vectors  Previous year questions (2016-20) - Motion Notes | EduRev have equal magnitudes. The magnitude of 

Previous year questions (2016-20) - Motion Notes | EduRev is n times the magnitude of Previous year questions (2016-20) - Motion Notes | EduRev The angle between Previous year questions (2016-20) - Motion Notes | EduRev is    (JEE Main 2019)
(1) Previous year questions (2016-20) - Motion Notes | EduRev 
(2) Previous year questions (2016-20) - Motion Notes | EduRev
(3) Previous year questions (2016-20) - Motion Notes | EduRev
(4) Previous year questions (2016-20) - Motion Notes | EduRev
Ans:
(1)
Solution:
We know that
Previous year questions (2016-20) - Motion Notes | EduRev
According to the question
Previous year questions (2016-20) - Motion Notes | EduRev
Squaring both the sides, we get
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev

Q 14. A body is projected at t = 0 with a velocity 10 m/s at an angle of 60° with the horizontal. The radius of curvature of its trajectory at t = 1 s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 m/s2, the value of R is     (JEE Main 2019)
(1) 2.5 m
(2) 2.8 m
(3) 10.3 m
(4) 5.1 m
Ans:
(2)
Solution:
At t = 1:
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
The radius of curvature of the trajectory is
Previous year questions (2016-20) - Motion Notes | EduRev
= 2.77 m = 2.8 m

Q 15. A particle moves from the point Previous year questions (2016-20) - Motion Notes | EduRev at t = 0 with an initial velocity Previous year questions (2016-20) - Motion Notes | EduRev It is acted upon by a constant force which produces a consult acceleration Previous year questions (2016-20) - Motion Notes | EduRev What is the distance of the particle from the origin at time 2s?    (JEE Main 2019)
(1) 15m
(2) 20√2m
(3) 5m
(4) 10√2m
Ans: 
(2)
Solution:
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev

Q 16. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.    (JEE Main 2018)

(1)

Previous year questions (2016-20) - Motion Notes | EduRev

(2)

Previous year questions (2016-20) - Motion Notes | EduRev

(3)

Previous year questions (2016-20) - Motion Notes | EduRev

(4)

Previous year questions (2016-20) - Motion Notes | EduRev

Ans: B
Solution:
In graph ‘2’ initial slope is zero which is not possible, since initial velocity is non-zero in all other three graphs.

Q 17. An automobile, travelling at 40 km/h, can be stopped at a distance of 40m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding):    (JEE Main 2018)
(1) 75 m
(2) 160 m
(3) 150 m
(4) 100 m
Ans:
(2)
Solution:

Previous year questions (2016-20) - Motion Notes | EduRev


Q 18. The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively.     (JEE Main 2018)
Previous year questions (2016-20) - Motion Notes | EduRev

(1) 337.5 m and 25 s
(2) 225.5 m and 10 s
(3) 112.5 m and 15 s
(4)  112.5 m and 22.5 s
Ans:
D
Solution:

The distance traveled in 15 seconds by both will be given by area under curve.
Previous year questions (2016-20) - Motion Notes | EduRev

Suppose after time t, they will meet, then distance traveled by both of them will be equal.
Previous year questions (2016-20) - Motion Notes | EduRev


Q 19. Let Previous year questions (2016-20) - Motion Notes | EduRev The magnitude of a coplanar vector Previous year questions (2016-20) - Motion Notes | EduRev is given by:     (JEE Main 2018)
(1) Previous year questions (2016-20) - Motion Notes | EduRev

(2) Previous year questions (2016-20) - Motion Notes | EduRev

(3) Previous year questions (2016-20) - Motion Notes | EduRev

(4)Previous year questions (2016-20) - Motion Notes | EduRev
Ans:
C
Solution:
Previous year questions (2016-20) - Motion Notes | EduRev
Solving equation (i) and (ii) we get
Previous year questions (2016-20) - Motion Notes | EduRev

Q 20. Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?    (JEE Main 2017)
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev
Ans:
(3)
Solution:
Previous year questions (2016-20) - Motion Notes | EduRev

Q 21. A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of (JEE Main 2017)
(1) Previous year questions (2016-20) - Motion Notes | EduRev
(2) 15 s
(3) Previous year questions (2016-20) - Motion Notes | EduRev
(4) Previous year questions (2016-20) - Motion Notes | EduRev
Ans: 
(4)
Solution:
Previous year questions (2016-20) - Motion Notes | EduRev
Previous year questions (2016-20) - Motion Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Previous year questions (2016-20) - Motion Notes | EduRev

,

Previous year questions (2016-20) - Motion Notes | EduRev

,

Extra Questions

,

ppt

,

practice quizzes

,

Previous Year Questions with Solutions

,

Free

,

Exam

,

Important questions

,

study material

,

pdf

,

Viva Questions

,

Summary

,

Previous year questions (2016-20) - Motion Notes | EduRev

,

Sample Paper

,

Semester Notes

,

shortcuts and tricks

,

video lectures

,

MCQs

,

mock tests for examination

,

past year papers

,

Objective type Questions

;