Question

A metal palate weighing 200 g is balance in mid air by throwing 40 balls per second vertically upwards from below. After collision, balls rebound with the same speed. What is the speed with which the balls strike the plate? [Given: mass of each ball = 200 g and $g=10m/s^2$]

• A. 5.7 cm/s
• B. 12.5 cm/s
• C. 9.8 cm/s
• D. 14 cm/s

Answer 1

The correct option is B 12.5 cm/s

We know that $F=\frac{\Delta p}{\Delta t}$
The plate balances in the air when the net force due to gravitation and the balls hitting the plate becomes equal.
For force due to balls hitting the plate, the change in momemtum will be
$\Delta p=nm(v-u)$
Since both initial and final velocity are same $⟹u=-v$
$⟹\Delta p=2nmv$
Since 40 balls hits per second,
$⟹$ for $\Delta t=1$ we have n=40
Then $F=\frac{2\times 40\times mv}{1}$
Gravitational force acting on the plate will be
$F=Mg$ where M is the mass of the plate
For the plate to balance both forces should be equal.
$\therefore 2\times 40\times mv=Mg$
$v=\frac{200\times 1000}{2\times 200\times 40}$
$=\frac{1000}{80}cm/s$
$=12.5cm/s$