JEE Main 2022 · Physics · Work, Energy and Power · medium

A 2 kg block moving at 6 m s^-1 on a smooth surface collides head-on with a stationary 4 kg block, the coefficient of restitution being 0.5.…

Numerical answer

2\ \mathrm{kg}

block moving at

6\ \mathrm{m\,s^{-1}}

on a smooth surface collides head-on with a stationary

4\ \mathrm{kg}

block, the coefficient of restitution being

0.5

. Immediately after the collision the

4\ \mathrm{kg}

block enters a rough region where the coefficient of kinetic friction is

0.25

. Taking

g=10\ \mathrm{m\,s^{-2}}

, the distance (in cm) travelled by the

4\ \mathrm{kg}

block in the rough region before stopping is

Show answer and solution

Answer: 180

Momentum:

2(6)=2v_1+4v_2

. Restitution:

v_2-v_1=0.5(6-0)=3

. Substituting

v_1=v_2-3

12=2v_2-6+4v_2

, so

v_2=3\ \mathrm{m\,s^{-1}}

and

v_1=0

. The

4\ \mathrm{kg}

block decelerates at

\mu g=2.5\ \mathrm{m\,s^{-2}}

, so

s=\dfrac{v_2^{2}}{2\mu g}=\dfrac{9}{5}=1.8\ \mathrm{m}=180\ \mathrm{cm}

. The mass of the block does not matter in the friction stage, but it does in finding its speed after the collision.

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