Question

A gun which fires small balls each of mass 20 gm is firing 20 balls per second on the smooth horizontal table surface ABCD If the collision is perfectly elastic and balls are striking at the centre of table with a speed 5 m/sec at an angle of ${}^{\circ }60$ with the vertical just before collision then force exerted by one of the leg on ground is (in N) (assume total weight of the table is 0.2 kg and g = 10 $m/s^2$)

Answer 1

As figure show forces acting on table is $=mg+F$

where $F=$force due to collision of ball.

F = $\frac{changeinmomentum}{time}$ or change in momentum in one second.

F = $\frac{n2n[V\sin \theta -(-V\sin \theta )]}{1}=20\times 2V\sin {30}^0\times 20\times {10}^{-3}$

F = $20\times 20\times 2\times 5\times \frac{n}{2}\times {10}^{-3}=2N$

Total downward force on table at center $=mg+f$

$F_{net}=0.2\times 10+2=2+2=4\sim$

$F_{net}$ is balance by all the four legs of table one leg exert force = $\frac{F_{net}}{4}=1N$