# A dish of mass 10g is kept horizontally in air by firing bullets of mass 5g each at the rate of 100/s. I the bullets rebound with the same speed,what is the velocity with which bullets are being fired?

## Take angle between bullets and dish is 90 degree.

Jan 20, 2017

The dish is being held horizontal in air by volley of bullets fired from below perpendicular to it. Therefore, in equilibrium position downward force due to gravity $\left(g = 9.81 {\text{ ms}}^{-} 2\right)$ is being balanced by equivalent upwards force generated by the bullets. Using SI system of units.
Downward force$= m g = \frac{10}{1000} \times 9.81 = 0.0981 \text{ N}$

We know that force $\vec{F}$ is the rate of change momentum $\vec{p}$
$\vec{F} = \frac{\mathrm{dv} e c p}{\mathrm{dt}}$

Let velocity of bullet as it collides the dish be$= v {\text{ ms}}^{-} 1$
Momentum of single bullet $= \frac{5}{1000} v = 0.005 v {\text{ kg ms}}^{-} 1$
The bullets collide with the dish elastically, momentum of one bullet after rebound$= - 0.005 v {\text{ kg ms}}^{-} 1$
Change of momentum of one bullet$= - 0.005 v - 0.005 v = - 0.01 v {\text{ kg ms}}^{-} 1$

By Law of Conservation momentum, equal and opposite to this momentum should be transferred to dish.

Setting up of force equation in the equilibrium position for bullets fired at the rate of $100 {\text{ s}}^{-} 1$we get

$0.0981 = 0.01 v \times 100$
$\implies v = \frac{0.0981}{0.01 \times 100}$
$\implies v = 0.0981 {\text{ ms}}^{-} 1$