# Question #1f1e2

Jan 19, 2017

Let a cricket ball of mass $m$ moving with velocity $v$ be stopped by a cricketer taking a catch within $\Delta t$ sec after the ball touching his hand. Here initial velocity is $v$ and final velocity is zero.So change in momentum during $\Delta t$ sec is $m v$.

By Newton's second law the average force $F$ exerted by the cricketer to stop the ball is equal to the rate of change of momentum $= \frac{m v}{\Delta t}$

So $\textcolor{b l u e}{F = \frac{m v}{\Delta t} \ldots . . \left(1\right)}$

Now let us see what the equation (1) tells us.

In this equation $m v$ is a constant for a particular ball of mass $m$ moving with a particular velocity $v$.

So force $F$ exerted by the cricketer to stop the ball is inversely proportional to $\Delta t$,the duration of application of retarding force $F$.

So while catching the fast moving cricket ball cricketer tries to lengthen the duration $\Delta t$ by withdrawing his hand towards the direction of motion of the ball in order to lessen the retarding force.
Because less retarding force creats less reaction on hand so that it gets less hurt and possibility of missing the catch for bouncing back from hand due to greater reactionary force is lessend.