Question
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?
Answers
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
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
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