Answer:
a) 4.583 m/s
b) 31.505 J
c) 0.491 m/s
d) 3.375 J
e)
p_player = (110 kg)(8 m/s) = 880 kg m/s
p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s
Explanation:
HI!
a)
We can calculate the recoil velocity by conservation of momentum, remember that p=mv.
The momentum of the bullet is:
p_b = (0.0250 kg)*(550 m/s )
The momentum of the rifle is:
p_r = (3 kg) * v
Since the total initial momentum is zero:
p_b = p_r
That is:
v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s
b)
The kinetic energy gained by the rifle is:
K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J
c)
We use the same formula as in a), but with m=28kg instead of 3 kg
v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s
d)
Again, the same formula as b, but with m=28 and v=0.491 m/s
K = 3.375 J
e)
p_player = (110 kg)(8 m/s) = 880 kg m/s
p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s
I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:
K = p^2/(2m)
usiing this relation, we get:
K_player = 3520 J
K_ball = 128.125 J
Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!
= 40°