Calculate the momentum of a bullet if its mass is 0.02 kg and it is moving with a
velocity of 70 m/s.
Answers
Mass of the bullet, m = 0.012 kg
Initial speed of the bullet, ub=70m/s
Mass of the wooden block, M=0.4 kg
Initial speed of the wooden block, uB =0
Final speed of the system of the bullet and the block = v m/s
Applying the law of conservation of momentum:
mub+MuB=(m+M)v
0.012×70+0.4×0=(0.012+0.4)v
v=0.84/0.412
=2.04 m/s
For the system of the bullet and the wooden block:
Mass of the system, m′=0.412 kg
Velocity of the system =2.04m/s
Height up to which the system rises = h
Applying the law of conservation of energy to this system:
Potential energy at the highest point = Kinetic energy at the lowest point
m′gh=(1/2)m′v2
h=2gv2
=2×9.8(2.04)2
=0.2123m
The wooden block will rise to a height of 0.2123m.
The heat produced = Kinetic energy of the bullet - Kinetic energy of the system
=(1/2)mu2−(1/2)m′v2
=(1/2)×0.012×(70)2−(1/2)×0.412×(2.04)2
=29.4−0.857=28.54J