Question

Consider a rifle, which has a mass of 2.44 kg and a bullet which has a mass of 150 grams and is loaded in the firing chamber.When the rifle is fired the bullet leaves the rifle with a muzzle velocity of 440 m/s;a. What will be the momentum of the rifle immediately before the bullet is fired?b. What will be the momentum of the rifle-bullet combination before the bullet is fired?c. What will be the momentum of the bullet immediately after the rifle has been fired?d. What will be the momentum of the rifle immediately after it has been fired?e. What will be the momentum of the rifle-bullet combination immediately after the bullet had been fired?f. What will be the velocity of the rifle immediately after the rifle has been fired?g. After the rifle has been fired it comes into contact with the marksman’s shoulder and then comes to a halt during a timeperiod of 0.38 seconds. What is the average force applied to the rifle by the shoulder?After the bullet leaves the rifle is strikes a block of wood which has a mass of 5.10 kg. and is sitting on a horizontalsurface which has a coefficient of sliding friction ofm= 0.83 .h. What will be the velocity of the bullet-block combination immediately after the bullet strikes the block of wood?i. How far will the block slide along the horizontal surface before it comes to a halt?j. How much energy was lost as the bullet was lodged in the block of wood?

Answer 1

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

$F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N$

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:

$W = K_e$

$dF_f = 0.5Mv_0^2$

$d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m$

j.Kinetic energy of the bullet before the impact:

$K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J$

Kinetic energy of the block-bullet system after the impact:

$K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J$

So 14520 - 33 = 14487 J was lost during the lodging process.