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

Hint: 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.