Q 1 :

If a rubber ball falls from a height h and rebounds up to the height of h/2. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are                 [2024]

  • 40%,2gh

     

  • 50%,gh2

     

  • 50%,2gh

     

  • 50%,gh

     

(3)  

       Velocity just before collision = 2gh

        Velocity just after collision = 2g(h2)

        ∴     ∆KE=12m(2gh)-12mgh=12mgh

        ∴     % loss in energy ∆KEKEi×100

         ⇒∆KEKEi×100=12mgh12mg2h×100=50%

 



Q 2 :

A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by (given g=9.8 m/s2)           [2024]

  • 4.32 kg ms-1

     

  • 43.2 kg ms-1

     

  • 23.9 kg ms-1

     

  • 2.39 kg ms-1

     

(4)

m=100g=0.1 kg, h1=10 m, h2=5 m

u=2gh1=2×98 m/s

v=2gh2=98 m/s

Impulse =ΔP→=P→f-P→i

=0.1[98+2×98]

≈2.39 kg m/s



Q 3 :

A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10-2 kg moving with a speed of 2×102 ms-1. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is (use g=10 m/s2)         [2024]

  • 0.30 m

     

  • 0.20 m

     

  • 0.35 m

     

  • 0.40 m

     

(2)

Pi=Pf⇒mu=(M+m)V

mu=(M+m)V

10-2×2×102≅1×V⇒V≅2 m/s

 



Q 4 :

As shown below, bob A of a pendulum having massless string of length 'R' is released from 60° to the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the centre. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (Take g as acceleration due to gravity)          [2025]

  • 13Rg

     

  • Rg

     

  • 43Rg

     

  • 23Rg

     

(1)

Velocity of a just before hitting:

u=2gR2=gR

Just after collision, let velocity of A and B are v1 and v2 respectively

By COM: mu=mv1+m2v2

2v1+v2=2u         ... (i)

e=1=v2–v1u

⇒ v2–v1=u          ... (ii)

From (i) and (ii)

⇒ 3v1=u ⇒ v1=u3=13gR



Q 5 :

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : 

Three identical spheres of same mass undergo one dimensional motion as shown in figure with initial velocities VA = 5 m/s, VB = 2 m/s, VC = 4 m/s. If we wait sufficiently long for elastic collision to happen, then VA = 4 m/s, VB = 2 m/s, VC = 5 m/s will be the final velocities.

Reason (R) : In an elastic collision between identical masses, two objects exchange their velocities.

In the light of the above statements, choose the correct answer from the options given below:           [2025]

  • Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

     

  • (A) is true but (R) is false

     

  • Both (A) and (R) are true and (R) is the correct explanation of (A)

     

  • (A) is false but (R) is true

     

(4)

In elastic collision for same mass, velocities interchange for

Final velocity

VA = 2 m/s

VB = 4 m/s

VC = 5 m/s



Q 6 :

As per the given figure, a small ball P slides down the quadrant of a circle and hits the other ball Q of equal mass which is initially at rest. Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball Q after collision will be (g = 10 m/s2)        [2023]

  • 0

     

  • 0.25 m/s

     

  • 2 m/s

     

  • 4 m/s

     

(3)

The velocities will be interchanged after collision. Speed of P just before collision =2gh

=2×10×0.2

=2 m/s



Q 7 :

A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travel independently. The ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g = 10 m/s2)            [2023]

  • 120 m/s

     

  • 60 m/s

     

  • 400 m/s

     

  • 360 m/s

     

(4)

v1=302hg,  v2=1202hg

(0.01)u=(0.2) 30g2h+(0.01) 120g2h

u=300+60=360 ms-1



Q 8 :

A bullet of mass 0.1 kg moving horizontally with speed 400 ms-1 hits a wooden block of mass 3.9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is ______.(Given g = 10 m/s2)    [2023]

  • 0.25

     

  • 0.50

     

  • 0.90

     

  • 0.65

     

(1)

Pi=Pf (collision)

⇒(0.1)(400)=(0.1+3.9)v

⇒v=0.1×4004=10 m/s

      a=μmgm=μg

Apply equation of motion, v2=u2+2as

⇒02=(10)2-2μg×20

⇒40μg=100

⇒μ=1002×10×20=14



Q 9 :

A particle of mass m moving with velocity v collides with a stationary particle of mass 2m. After collision, they stick together and continue to move together with velocity ______.         [2023]

  • v

     

  • v2

     

  • v3

     

  • v4

     

(3)

Applying conservation of linear momentum

⇒P→i=P→f

mv+2m×0=(3m)v'

∴   mv=3mv' ⇒ v'=v3



Q 10 :

A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 m/s. The initial speed of the smaller body before collision is ________ ms-1.           [2023]



(4)

1×u1=-2+3v

⇒u1=-2+3v                       ...(i)

1=v+2u1 ⇒ v+2=u1         ...(ii)

Solving (i) and (ii), u1=4 m/s



Q 11 :

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height of 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ________ ms-2. [Given g=10 ms-2]               [2023]



(120)

vi=2ghi=2×10×9.8 ↓=14 m/s↓

vf=2ghf=2×10×5 ↑=10 m/s↑

|a→avg|=|Δv→Δt|=240.2=120 m/s2



Q 12 :

A ball is dropped from a height of 20 m. If the coefficient of restitution for the collision between ball and floor is 0.5, after hitting the floor, the ball rebounds to a height of ________ m.          [2023]



(5)

We know h'=e2h

h'=(0.5)2×20 m=5 m



Q 13 :

A small bob A of mass m is attached to a massless rigid rod of length 1 m pivoted at point P and kept at an angle of 60° with the vertical as shown in the figure. At a distance of 1 m below point P, an identical bob B is kept at rest on a smooth horizontal surface that extends to a circular track of radius R as shown in the figure. If bob B just manages to complete the circular path of radius R up to a point Q after being hit elastically by bob A, then the radius R is ______ m.              [2026]

  • 15

     

  • 35

     

  • 2+35

     

  • 2-35

     

(1)

VA at lowest point

VA=2gl(1-cosθ)

VA=2×10×1(1-12)=10

After collision velocity of B becomes,

VA=10=VB  (Same mass)

Now to complete circular motion

VB=5gR

R=15



Q 14 :

Assertion (A): A body of mass m1 collides head on elastically with another body of mass m2 at rest. The ratio of the final energy of the first body to the final energy of the second body is (m1−m2)24m1m2

Reason (R): The collision is perfectly elastic and the coefficient of restitution is 1.

  • Both (A) and (R) are true and (R) is the correct explanation of (A)

     

  • Both (A) & (R) are true but (R) is not correct explanation of (A)

     

  • (A) is true and (R) is false

     

  • Both (A) and (R) are false

     

(1)

K.E.1=12m1v12;    K.E.2=12m2v22

K.E.1K.E.2=12(m1-m2m1+m2)2u1212m2(2m1m1+m2)2u12=(m1-m2)24m1m2

For perfectly elastic collision, e=1



Q 15 :

A moving particle of mass m, makes a head on elastic collision with another particle of mass 2m, which is initially at rest. The percentage loss in energy of the colliding particle on collision, is close to ..........



(89)

Fractional decrease in kinetic energy of mass 'm'

=1-(m2-m1m2+m1)2=1-(2-12+1)2=1-(13)2=1-19=89

Percentage loss in energy =89×100≈89%



Q 16 :

Statement-1: Two particle moving in the same direction do not lose all their energy in a completely inelastic collision.

Statement-2: In accordance with the principle of conservation of momentum, kinetic energy will not be zero once collision is over.

  • Statement-1 is true, Statement-2 is false

     

  • Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1

     

  • Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1

     

  • Statement-1 is false, Statement-2 is true

     

(2)

Conceptual



Q 17 :

Two particles of equal mass m have respective initial velocities ui^ and u(i^+j^2).  They collide completely inelastically. The energy lost in the process is

  • 34mu2

     

  • 23mu2

     

  • 13mu2

     

  • 18mu2

     

(4)

Ki=12m(u2+u22)=34mu2

Kf=12(2m)u216×10=58mu2

Loss in kinetic energy =34mu2-58mu2=18mu2



Q 18 :

A particle of mass m moves towards a smooth vertical wall with a speed u (relative to the ground) and collides elastically with the wall; the wall moving towards the particle with a speed 2u (also relative to the ground). Assuming that the wall is extremely massive, the magnitude of impulse delivered to the particle equals

  • 5 mu

     

  • 6 mu

     

  • 4 mu

     

  • 2 mu

     

(2)

In 1D collisions formula for v1=m1-m2m1+m2v1+2m2m1+m2v2, when m1≪m2, v1→=-u1→+2u2→.