Consider a horizontal surface moving vertically upward with velocity $2 m s^{- 1}$ . A small ball of mass $2 kg$ is moving with velocity $(2 \hat{i} - 2 \hat{j}) m s^{- 1}$ If the coefficient of restitution and coefficient of friction are $\frac{1}{2}$ and $\frac{1}{3}$ respectively, find the horizontal velocity (in $m s^{- 1}$ ) of the ball after the collision.

Question

Consider a horizontal surface moving vertically upward with velocity

2 m s^{- 1}

. A small ball of mass

2 kg

is moving with velocity

(2 \hat{i} - 2 \hat{j}) m s^{- 1}

If the coefficient of restitution and coefficient of friction are

\frac{1}{2}

and

\frac{1}{3}

respectively, find the horizontal velocity (in

m s^{- 1}

) of the ball after the collision.

Aicharya · Accepted Answer

The relative velocity of the particle with respect to the surface, along the normal, is the approach speed of the particle.

$v_{app} = 4 m s^{- 1}$

The separation velocity will be

$v_{sep} = e v_{app} = \frac{1}{2} \times 4 = 2 m s^{- 1}$

$∴ {\bar{v}}_{m, g r} = {\bar{v}}_{m, s} + {\bar{v}}_{s, g r} = 4 m/s$

$\int N d t = 4 (2) + 2 (2) = 12$

$Now, \int μ N \cdot d t = 2 (v_{x f} - 2)$

$= - \frac{1}{3} \times 12 = 2 (v_{x f} - 2)$

${(v_{x})}_{final} = 0$

Consider a horizontal surface moving vertically upward with velocity 2 m s - 1 . A small ball of mass 2 kg…