Question

(a) Use Newton’s second law of motion to explain the following:

(i) A cricketer pulls his hands back while catching a fast moving cricket ball.

(ii) An athlete prefers to land on sand instead of hard floor while taking a high jump.

(b) A ball is thrown vertically upwards. It goes to a height of $19.6m$ and then comes back to the ground. Find :

(i) the initial velocity of the ball,

(ii) the total time of journey, and

(iii) the final velocity of the ball when it strikes the ground.

Take $g=9.8m{s}^{-2}$.

Answer 1

(a)

(i) We pull our hands back while catching a fast moving cricket ball, because by doing so, we increase the time of catch, i.e. increase the time to bring about a given change in momentum, and hence, the rate of change of momentum decreases. Thus, a small force is exerted on our hands by the ball.

(ii) When an athlete lands from a height on a hard floor, his feet comes to rest instantaneously, so a very large force is exerted by the floor on his feet, but if he lands on sand, his feet push the sand for some distance; therefore, the time duration in which his feet comes to rest increases. As a result, the force exerted on his feet decreases and he is saved from getting hurt.

(b) Given, $h=19.6m,a=-g=-9.8m{s}^{-2},v=0$

(i) From equation of motion $v^2=u^2+2gh$
$$0=u^2-2\times 9.8\times 19.6$$
$$u^2=19.6\times 19.6$$
$\therefore$ Initial velocity $u=19.6m{s}^{-1}$.

(ii) Let $ts$ be the time taken by the ball to reach the highest point.

From equation of motion $v=u+gt$
$$0=19.6-9.8t\text{or}9.8t=19.6$$
$\therefore t=\frac{19.6}{9.8}=2s$
It will take the same time $t=2s$ to come back from the highest point to the ground.
$\therefore$ Total time of journey $t^{\prime }=2t=2\times 2=4s$.

(iii) The final velocity of ball when it strikes the ground will be same as the initial velocity with which it was thrown upwards.

Final velocity on reaching the ground $=19.6m{s}^{-1}$.