Question

A ball of mass 400 g is dropped from a height of 5 m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 N so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the bat is $[g=10m/s^2]$

• A. 0.12 s
• B. 0.08 s
• C. 0.04 s
• D. 12 s

Answer 1

The correct option is A 0.12 s

$v^2–u^2=2as$
$u=0$ since ball is dropped
$v=\sqrt{2as}$
$v_i=\sqrt{2\times 10\times 5}m/s=10m/s$
$p_i=m{v}_i$
$=0.4kg\times 10m/s$
$=4kgm/s$
$p_i$ is directed downward.
$v_f=\sqrt{2\times 10\times 20}m/s=20m/s$
$P_f=0.4kg\times 20m/s=8kgm/s$
$p_f$ is directed vertically upwards.
Now, impluse = change of linear momentum$\Rightarrow (–100)\times \Delta t=(–8–4)$
[Force and velocity are directed upwards]
[Giving negative sign to the vertically upwards direction]
$\Delta t=\frac{12}{100}=0.12s$