A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. (a) Calculate the recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? (d) How much kinetic energy is transferred to the rifleshoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation. (e) Calculate the momentum of a 110-kg football player running at 8.00 m/s. Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Discuss its relationship to this problem.

(a) 4.58 m/s

(b) 31.5 J

(c) 0.491 m/s

(d) 0.362 J

(e) Momentum of player = 880 kg m/s

Momentum of ball = 10.3 kg m/s

Explanation:

(a) By Newton's third law of motion, action and reaction are equal and opposite.

The momentum action of bullet = (0.0250 kg)(550 m/s)

The momentum reaction of rifle = (3.00 kg)(v m/s)

Hence,

(0.0250 kg)(550 m/s) = (3.00 kg)(v m/s)

v =(0.0250 kg)(550 m/s)/(3.00 kg) = 4.58 m/s

(b) Kinetic energy =

(c) If the effective mass of the rifle is now 28.0 kg, then it's recoil speed is

v =(0.0250 kg)(550 m/s)/(28.0 kg) = 0.491 m/s

(d) With the new effective mass, kinetic energy is

(e) The momentum of the player =(110 kg)(8.00 m/s) = 880 kg m/s

Momentum of ball = (0.41 kg)(25.0 m/s) = 10.3 kg m/s

The momentum of the bullet in the question is = (0.0250 kg)(550 m/s) = 13.8 kg m/s.

This momentum is much smaller than that of the player. Hence, an object or body will feel more 'pain' or impact when hit by the running player than by the bullet.

The impact of the bullet is only slightly more than that of the thrown ball.

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