Question

A ball is dropped on the floor from height of $10\,{\text{m}}$. It rebounds to a height of $2.5\,{\text{m}}$. If the ball is in contact with the floor for $0.01\,{\text{sec}}$, then average acceleration during the contact is
A. $2100\,{\text{m/}}{{\text{s}}^2}$
B. $1400\,{\text{m/}}{{\text{s}}^2}$
C. $700\,{\text{m/}}{{\text{s}}^2}$
D. $400\,{\text{m/}}{{\text{s}}^2}$

Answer 1

Use the kinematic equation relating final velocity, initial velocity, acceleration and time of the ball. Also, use the acceleration of the ball during the contact time.

Formulae used:

The kinematic equation relating final velocity $v$, initial velocity $u$, acceleration $g$ and displacement $h$ of particle in a free fall is given by
${v^2} = {u^2} - 2gh$ …… (1)

The acceleration $a$ of a particle is given by
$a = \dfrac{{\Delta v}}{{\Delta t}}$ …… (2)

Here, $\Delta v$ is the change in velocity of the particle in the time interval $\Delta t$.

Complete step by step answer: When the ball is first time dropped from the height ${h_1}$ of $10\,{\text{m}}$, its initial velocity ${u_1}$ is zero and the final velocity is ${v_1}$.
${u_1} = 0\,{\text{m/s}}$

Calculate the final velocity of the ball when it first time hits the floor.

Rewrite equation (1) for the final velocity ${v_1}$ of the ball.
$v_1^2 = u_1^2 - 2g{h_1}$

Substitute $0\,{\text{m/s}}$ for ${u_1}$, $9.8\,{\text{m/}}{{\text{s}}^2}$ for $g$ and $10\,{\text{m}}$ for ${h_1}$ in the above equation.
$v_1^2 = {\left( {0\,{\text{m/s}}} \right)^2} - 2\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {10\,{\text{m}}} \right)$
$\Rightarrow v_1^2 = - 196$

Take square root on both sides.
$\Rightarrow {v_1} = - 14\,{\text{m/s}}$

Hence, the velocity of the ball when it first strikes the floor is $- 14\,{\text{m/s}}$.

When the particle hits the floor first time, its displacement ${h_2}$ is $2.5\,{\text{m}}$ in the vertical direction. The initial velocity ${u_2}$ of this ball is and the final velocity ${v_2}$ is zero.

Rewrite equation (1) for the final velocity ${v_2}$ of the ball.
$v_2^2 = u_2^2 - 2g{h_2}$

Substitute $0\,{\text{m/s}}$ for ${v_2}$, $9.8\,{\text{m/}}{{\text{s}}^2}$ for $g$ and $2.5\,{\text{m}}$ for ${h_2}$ in the above equation.
${\left( {0\,{\text{m/s}}} \right)^2} = u_2^2 - 2\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {2.5\,{\text{m}}} \right)$
$\Rightarrow u_2^2 = 49$

Take square root on both sides of the above equation.
${u_2} = 7\,{\text{m/s}}$

Hence, the velocity of the ball after it strikes the floor is $7\,{\text{m/s}}$ in the upward direction.

Calculate the acceleration of the ball during the time interval $0.01\,{\text{sec}}$.

Substitute ${u_2} - {v_1}$ for in equation (2).
$a = \dfrac{{{v_1} - {u_2}}}{{\Delta t}}$

Substitute $- 14\,{\text{m/s}}$ for ${v_1}$, $7\,{\text{m/s}}$ for ${u_2}$ and $0.01\,{\text{s}}$ for $\Delta t$ in the above equation.
$a = \dfrac{{\left( {7\,{\text{m/s}}} \right) - \left( { - 14\,{\text{m/s}}} \right)}}{{0.01\,{\text{s}}}}$
$\Rightarrow a = \dfrac{{21\,{\text{m/s}}}}{{0.01\,{\text{s}}}}$
$\Rightarrow a = 2100\,{\text{m/}}{{\text{s}}^2}$

Therefore, the acceleration of the ball is $2100\,{\text{m/}}{{\text{s}}^2}$.

Hence, the correct option is A.

Note: The sign of the velocity ${v_1}$ is negative because the direction of the velocity is in the downward direction.