Question

A cricket ball of mass $0.15kg$ is moving with a velocity of $12m{s}^{-1}$ and is hit by a bat so that the ball is turned back with a velocity of $20m{s}^{-1}$. If the force of blow acts for $0.01s$, find the average force exerted on the ball by the bat.

Answer 1

Step1: Given data

A cricket ball of mass $m=0.15kg$

The initial velocity of ball, $u=12m{s}^{-1}$

Final velocity of $v=20m{s}^{-1}$.

The force of blow acts for

$t=0.01s$

Step2: Formula used

$F=ma[F=force,m=mass,a=acceleration]$

Step3: Calculating velocity and acceleration

According to the given data, $u=-12m{s}^{-1}$ (-ve sign is taken because initial and final velocities are in opposite directions) and $v=20m{s}^{-1}$

$$\begin{gathered} changeinvelocity=v-u \\ changeinvelocity=20-(-12) \\ changeinvelocity=20+12 \\ changeinvelocity=32m{s}^{-1} \end{gathered}$$

Average acceleration for an interval is defined as the change in velocity for that interval per time.

$$\begin{gathered} a=\frac{changeinvelocity}{time} \\ a=\frac{32}{0.01}=3200m{s}^{-2} \end{gathered}$$

Step4: Calculating force

According to Newton's second law, force is stated as the "product of mass and acceleration of a body".

$$\begin{gathered} F=m\times a \\ F=0.15\times 3200 \\ F=480N \end{gathered}$$

Hence, the average force exerted on the ball by the bat is $480N$.