A pistol of mass 2 kg fires a bullet of mass 50g. The bullet strikes a stationary block of mass ½ kg. if the block, with the bullet in it, moves with a velocity of 4 m/s the recoil velocity of the pistol will be?

Answer 1

mass of block with bullet = .5 + .050 = .55 kg

momentum of block with bullet = .55*4
= 2.2 kg m/s

so momentum of bullet before it hits = 2.2

so .050 Vbullet = 2.2
Vbullet = 44m/s

but before we shot, the momentum was zero

0 = mass pistol * 0 + mass bullet * 0
no external foceres so after the shout momementum still 0

0 = 2 Vgun + momentum of bullet
= = 2 Vgun + 2.2
Vgun = - 1.1 m/s