Answer:
(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 = [tex]\frac{1}{2}mv^2 = \frac{1}{2}\times3.00\text{ kg}\times (4.58\text{ m/s})^2 = 31.5\text{ J}[/tex]
(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
[tex]\frac{1}{2}mv^2 = \frac{1}{2}\times3.00\text{ kg}\times (0.491\text{ m/s})^2 = 0.362\text{ J}[/tex]
(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.
