A3.5-kg rifle fires a 15-g bullet at 840 m/s . with what speed does the rifle recoil (move backward toward the shooter's shoulder)?

Momentum before the action = Momentum after

So:

Where m1 and v1 are for the gun and m2 and v2 are for the bullet

(initial side) m1v1 + m2v2 = m1v1 + m2v2 (final side)

Initially there is no momentum so we are left with

0 = m1v1 + m2v2 (final)

Subtract over m1v1

-m1v1 = m2v2

We are looking for v1 which is the final velocity of the gun

so:

v1 = m2v2 / -m1

v1 = (15 x 10^-3 kg)(840) / -(3.5)

v1 = -3.6 m/s

The negative indicates that it is travelling in the negative direction towards the shooter

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j