Question

A cricket ball of mass $0.25Kg$ with speed $10 m/s$ collides with a bat and returns with same speed within \[0.01s\]. The force acted on bat is
(A). \[25\;N\]
(B). \[50\;N\]
(C). \[250\;N\]
(D). \[500\;N\]

Answer 1

Hint- In such types of cases the role of the momentum comes to calculate the force on the ball. The force on the ball which is being acted during the collision with the bat and returning back is the change in momentum divided by the change in the time.

Formula Used:
$F = \dfrac{{\Delta p}}{{\Delta t}}$-----equation (1)
where, $\Delta p = $change in momentum
$\Delta t = $change in time.

Complete step-by-step answer:
The given quantities in the question,
Mass of the cricket ball= $m = 0.25Kg$
Initial Speed of the cricket ball=$u = 10 m/s$
Final speed of the cricket ball =$v = 10 m/s$
Change in time in which the ball returns with the same speed $ = \Delta t = 0.01s$
Change in the momentum of the ball =$\Delta p = mv - mu$
$ \Rightarrow \Delta p = (0.25 \times 10) - 0.25 \times \left( { - 10} \right)$
$ \Rightarrow \Delta p = 5Kg m/s$
Now we have calculated the momentum and the change in the time is already given. So, by using the equation (1) we will calculate the force acting on the bat.
Force acting on the bat=$F = \dfrac{{\Delta p}}{{\Delta t}}$
$ \Rightarrow F = \dfrac{5}{{0.01}}$
$ \Rightarrow F = 500N$
Hence the force acting on the bat is $500N$.
So, the option (D) is the correct answer.

Note- We know that Newton's second law of motion says that the acceleration of any object which is under a net force, is directly proportional to the magnitude of the net force, and inversely proportional to the mass of that object. In terms of the momentum the Newton’s second law of motion can be stated in the terms of the momentum and the time in which the momentum of the body changed. So, Newton's second law of motion (in terms of the momentum) states that the net external force equals the change in momentum divided by the change in the time over which it changes.