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
[tex]F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N[/tex]
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:
[tex]W = K_e[/tex]
[tex]dF_f = 0.5Mv_0^2 [/tex]
[tex]d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m[/tex]
j.Kinetic energy of the bullet before the impact:
[tex]K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J[/tex]
Kinetic energy of the block-bullet system after the impact:
[tex]K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J[/tex]
So 14520 - 33 = 14487 J was lost during the lodging process.
