Question

A pistol of mass 2 kg fires a bullet of mass 50 g. The bullet strikes a stationary block of mass 1/2 kg. If the block, with the bullet embedded in it, moves with a velocity of 4 m/s, then what will be the recoil velocity of the pistol ? (Ignore friction and air resistance)

Answer 1

Initially a pistol fires a bullet and recoils:

let Mp is mass of pistol and Mb is mass of bullet, let the recoil vel of pistol be vp and velocity of bullet is vb;

then by conservation of momentum

-Mp. vp + Mb.vb =0

vp = (Mb/Mp).vb = (50. 10^-3) / 2) .vb

so to find velocity of the bullet , the second event is given

the bullet attaches to a plank at rest and then both moves with 4m/s

Initial momentum is

Mb.vb =final momentum = {M(plank) + Mb }. 4m/s

therefore vb ={ (0.5 + 0.05) .4 }/ 0.05 = (0.55/0.05) . 4 m/s= 44m/s

so the bullet velocity is 44 m/s

therefore using above expression for recoil speed

vp = 25.10^-3 . 44 m/s = 1.1 m/s

the pistol should recoil with a speed 1.1 m/s