Question

A ball of mass 50 g is dropped from a height of 20 m.A boy on the ground hits the ball vertically upwards with a bat with an average force of 200 N, so that it attains a vertical height of 45 m. The time for which the ball remains in contact with the bat is [Take $g=10m/{{s}^{2}}$ ]

A. 1/20 of a second
B. 1/40 of a second
C. 1/80 of a second
D.1/20 of a second

Answer 1

Hint: At first we need to find the velocity acquired by the ball when it is falling down, then we need to find the velocity acquired when the ball is hit by the bat. Then we need to calculate the change in velocity of the ball. With this value we can find the change in momentum of the ball. Now we know that momentum has a relation with time and force using this relation we need to find the time.

Formula Used:
${{V}^{2}}={{U}^{2}}+2gs$
$\vartriangle t=\dfrac{\vartriangle p}{f}$
$m{{V}_{2}}-m{{V}_{1}}=m\vartriangle v$
$\vartriangle v={{V}_{2}}-{{V}_{1}}$

Complete step by step answer:
We are given that a ball has a mass of 50g.
It is dropped from a height of 20m.
Now, a boy on the ground hits the ball at a force of 200N vertically upwards, in such a way that it reaches a height of 45m.
So, let us consider the velocity of the ball just before hitting as ${{V}_{1}}$
Now if we want to find the value of ${{V}_{1}}$, we need to apply equation of motion,
${{V}^{2}}={{U}^{2}}+2gs$,
Now, as the ball is dropped, U = 0 m/s.
And we know that S = 20 m.
So,
${{V}_{1}}^{2}=0+2\times 10\times 20$ (We are putting the values in the equation of motion),
Now on solving this we get,
${{V}_{1}}=-20m/s$, this is negative as the ball is falling downwards.
Now, the height attained by the ball after being hit is 45m.
So, let us considered that the ball is hit by the bat with a velocity ${{V}_{2}}$
Now,
Again applying values in the equation of motion,
${{V}^{2}}={{U}^{2}}+2gs$
${{U}^{2}}={{V}^{2}}-2gs$
$0={{V}_{2}}^{2}-2\times 45\times 10$, We have considered U as zero because when the ball is hit there a time comes when the ball has no velocity that is the ball is stationary for an infinitesimal small time before attaining a velocity again.
On solving the above equation we get,
${{V}_{2}}=30m/s$, Now it is travelling upwards so it is positive.
So now the change in velocity is,
$\vartriangle v={{V}_{2}}-{{V}_{1}}$
$\vartriangle v=30-(-20)=50m/s$.
Now the change in momentum is, $m{{V}_{2}}-m{{V}_{1}}=m\vartriangle v=\dfrac{50}{1000}\times 50=2.5kg\text{ }m/s$.
So we know that force (f)= change in momentum / change in time.
So,
Change in time = force / change in momentum.
$\vartriangle t=\dfrac{\vartriangle p}{f}$,
$\vartriangle t=\dfrac{2.5}{200}=\dfrac{1}{80}\sec $.
So we can say that option C is the correct option.

Note:
We are using ${{V}^{2}}={{U}^{2}}+2gs$, instead of ${{V}^{2}}={{U}^{2}}+2as$, as the acceleration due to gravity is only acting on the ball. Here in this equation ‘V’ is the final velocity and ‘U’ is the initial velocity and ‘S’ is the distance attained. Students must carefully solve the question as it contains many parts of solving one may cause error in any of the parts.