A revolver of mass 500g fires a bullet of mass 10g with a speed of 100m/s. Find:
(i) The momentum of the bullet.
(ii) Initial momentum of revolver and bullet as a whole.
(iii) Recoil velocity of the revolver
These questions are NOT clubbed.
(i) The momentum of the bullet.
(ii) Initial momentum of revolver and bullet as a whole.
(iii) Recoil velocity of the revolver
Dear student,
1) Initial momentum of bullet p = mb x vb = (0.01) Kg x (100) m/s = 1 Kg m/s
2) Since initially the velocity of both the revolver and the bullet is zero so
p(initial) = mrvr + mbvb = 0.5 x 0 + 0.01 + 0 = 0 Kg m/s
3) Initial momentum = Final momentum
Initial momentum is 0 => Final momentum = 0
p(final) = mrvr + mbvb = 0.5v + 0.01 x 100 = 0
0.5 v + 1 = 0
So v = -1/0.5 = 2 m/s
Thus recoil velocity of gun is 2 m/s. It is in opposite direction to the velocity of bullet.
Regards
1) Initial momentum of bullet p = mb x vb = (0.01) Kg x (100) m/s = 1 Kg m/s
2) Since initially the velocity of both the revolver and the bullet is zero so
p(initial) = mrvr + mbvb = 0.5 x 0 + 0.01 + 0 = 0 Kg m/s
3) Initial momentum = Final momentum
Initial momentum is 0 => Final momentum = 0
p(final) = mrvr + mbvb = 0.5v + 0.01 x 100 = 0
0.5 v + 1 = 0
So v = -1/0.5 = 2 m/s
Thus recoil velocity of gun is 2 m/s. It is in opposite direction to the velocity of bullet.
Regards