Answer:
a. 0 kgm/s
b. 0 kgm/s
c. 66 kgm/s
d. -66 kgm/s
e. 0 kgm/s
f. -27.05 m/s
g. 173.68 N
h. 12.58 m/s
i. 0.772 m
j. 14487 J
Explanation:
150 g = 0.15 kg
a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.
b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.
c. After the bullet is fired, the momentum is:
0.15*440 = 66 kgm/s
d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s
e. 0 according to law of momentum conservation.
f. Velocity of the rifle is its momentum divided by mass
v = -66 / 2.44 = -27.05 m/s
g. The average force would be the momentum divided by the time
f = -66 / 0.38 = 173.68 N
h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is
66 / 5.25 = 12.58 m/s
i. The normal force and also friction force due to sliding is
According to law of energy conservation, the initial kinetic energy will soon be transformed to work done by this friction force along a distance d:
j.Kinetic energy of the bullet before the impact:
Kinetic energy of the block-bullet system after the impact:
So 14520 - 33 = 14487 J was lost during the lodging process.
