Question

7. 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 time

period 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 horizontal

surface which has a coefficient of sliding friction of

m

= 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

Solution) M = 2.44 kg

m = 150 g = 0.150 kg

v = 440 m/s

(a) Momentum of rifle(Pr) immediately before bullet is fired , Pr = ?

Pr = MV

V = 0

Pr = 2.44×0 = 0

Pr = 0 kgm/s

(b) Momentum of rifle - bullet (Prb) system before bullet is fired ,

Prb = ?

Prb = (m+M)V = (m+M)(0) = 0

Prb = 0 kgm/s

(c) Momentum of bullet after rifle has been fired , Pb = ?

Pb = mv = 0.150×440 = 66 kgm/s

Pb = 66 kgm/s

(d) Momentum of rifle after it has been fired , Prf = ?

Pb + Prf = 0

Prf = - Pb = - 66

Prf = - 66 kgm/s

* kindly post next question in next post

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