Question

A ball of mass 400 gm is dropped from a height of 5 m. A boy on the ground hits the ball vertically upward with a bat with an average force of 100 N so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the ball is. (take \[g = 10\,m/{s^2}\])
(A) \[0.12s\]
(B) \[0.08s\]
(C) \[0.04s\]
(D) \[12s\]

Answer 1

Hint: Use kinematical equation relating final velocity, initial velocity, acceleration and displacement to calculate the velocity of the ball just before it hits the bat and just after it hits the bat. Calculate the impulse in this process and then use the formula for impulse to determine the time of contact of the ball with the bat.

Formula used:
\[{v^2} = {u^2} + 2as\]
Here, v is the final velocity, u is the initial velocity, a is the acceleration and t is time.

Complete step by step answer:
Use the third kinematic equation relating initial velocity, final velocity, displacement, and acceleration to determine the final velocity of the ball just before it hits the bat as follows,
\[v_1^2 = u_1^2 + 2g{h_1}\]
Here, \[{v_1}\] is the final velocity of the ball just before it hits the bat, \[{u_1}\] is the initial velocity of the ball, g is the acceleration due to gravity and \[{h_1}\] is the initial height.
Since the initial velocity of the ball is zero, the above equation becomes,
\[v_1^2 = 2g{h_1}\]
\[ \Rightarrow {v_1} = \sqrt {2g{h_1}} \]
Substitute \[g = 10\,m/{s^2}\] and \[{h_1} = 5\,m\] in the above equation.
\[{v_1} = \sqrt {2\left( {10\,m/{s^2}} \right)\left( {5\,m} \right)} \]
\[ \Rightarrow {v_1} = \sqrt {100\,{m^2}/{s^2}} \]
\[\therefore {v_1} = 10\,m/s\]
Since the ball is accelerated downwards, the final velocity of the ball \[{v_1}\] is negative. Therefore,
\[{v_1} = - 10\,m/s\]
Again, use the kinematic relation to determine the initial velocity of the ball just after it is hit by the bat as follows,
\[v_2^2 = u_2^2 + 2g{h_2}\]
Here, \[{v_2}\] is the final velocity of the ball at maximum height, \[{u_2}\] is the initial velocity of the ball just before it is hit by the bat and \[{h_2}\] is the maximum height attained by the ball.
At maximum height, the final velocity of the ball becomes zero. Therefore, the above equation becomes,
\[u_2^2 + 2g{h_2} = 0\]
\[ \Rightarrow u_2^2 = - 2g{h_2}\]
\[\therefore {u_2} = \sqrt { - 2g{h_2}} \]
For the upward projection of the object, the acceleration due to gravity is negative. Substitute \[g = - 10\,m/{s^2}\] and \[{h_2} = 20\,m\] in the above equation.
\[{u_2} = \sqrt { - 2\left( { - 10\,m/{s^2}} \right)\left( {20\,m} \right)} \]
\[ \Rightarrow {u_2} = \sqrt {400\,{m^2}/{s^2}} \]
\[\therefore {u_2} = 20\,m/s\]
In this process, the change in the momentum of the ball is,
\[\Delta P = I = m{u_2} - m{v_1}\]
\[ \Rightarrow I = m\left( {{u_2} - {v_1}} \right)\]
Here, \[I\] is the impulse.
Substitute \[0.4\,kg\] for m, \[20\,m/s\] for \[{u_2}\] and \[ - 10\,m/s\] for \[{v_1}\] in the above equation.
\[
I = \left( {0.4\,kg} \right)\left( {\left( {20\,m/s} \right) - \left( { - 10\,m/s} \right)} \right) \\
\Rightarrow I = 12\,kg\,m/s \\
\]
The impulse is the product of force and time of contact of the force.
\[I = Ft\]
\[ \Rightarrow t = \dfrac{I}{F}\]
Substitute \[12\,kg\,m/s\] for I and 100 N for F in the above equation.
\[t = \dfrac{{12\,kg\,m/s}}{{100\,N}}\]
\[\therefore t = 0.12\,s\]
So, the correct answer is option (A).

Note: In this question, the downward direction of the ball is taken as negative and the vertical direction is taken as positive. Therefore, the velocity of the ball just before it hits the ball is \[{v_1} = - 10\,m/s\].