(a) -4.6 m/s
We can solve this part by using the law of conservation of momentum: in fact, the total momentum of the bullet-rifle system before and after the shot must be equal.
Before the shot, the total momentum is zero:
p = 0 (1)
After the shot, the total momentum is:
(2)
where
m = 0.0250 kg is the mass of the bullet
v = 550 m/s is the velocity of the bullet
M = 3.00 kg is the mass of the rifle
V is the recoil velocity of the rifle
Since momentum is conserved, (1) = (2), so we can solve for V:
![0=mv+MV\\V=-\frac{mv}{M}=-\frac{(0.0250)(550)}{3.00}=-4.6 m/s](https://tex.z-dn.net/?f=0%3Dmv%2BMV%5C%5CV%3D-%5Cfrac%7Bmv%7D%7BM%7D%3D-%5Cfrac%7B%280.0250%29%28550%29%7D%7B3.00%7D%3D-4.6%20m%2Fs)
And the negative sign means the rifle will move backward.
(b) 31.7 J
The kinetic energy of an object is given by
![K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where
m is the mass of the object
v is its speed
The rifle has a mass of
M = 3.00 kg
And a final speed of (speed = magnitude of velocity)
V = 4.6 m/s
Therefore, the kinetic energy it has gained is
![K=\frac{1}{2}(3.00)(4.6)^2=31.7 J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%283.00%29%284.6%29%5E2%3D31.7%20J)
(c) -0.5 m/s
In this case, we just need to repeat the problem as in part (a), applying the law of conservation of momentum:
![0=mv+MV](https://tex.z-dn.net/?f=0%3Dmv%2BMV)
where in this case, the mass of the rifle is
M = 28.0 kg
while the other data are unchanged:
m = 0.0250 kg is the mass of the bullet
v = 550 m/s is the velocity of the bullet
Solving for V, we find the new recoil velocity of the rifle:
![V=-\frac{mv}{M}=-\frac{(0.0250)(550)}{28.0}=-0.5 m/s](https://tex.z-dn.net/?f=V%3D-%5Cfrac%7Bmv%7D%7BM%7D%3D-%5Cfrac%7B%280.0250%29%28550%29%7D%7B28.0%7D%3D-0.5%20m%2Fs)
(d) 3.5 J
As in part b), we can apply the equation of the kinetic energy:
The kinetic energy of an object is given by
![K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where in this case, we have:
M = 28.0 kg is the mass of the rifle+shoulder
V = 0.5 m/s is the recoil speed of the rifle+shoulder
Substituting into the equation,
![K=\frac{1}{2}(28.0)(0.5)^2=3.5 J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%2828.0%29%280.5%29%5E2%3D3.5%20J)
(e) Player's momentum is larger
The momentum of the player is
![p'=MV](https://tex.z-dn.net/?f=p%27%3DMV)
where
M = 110 kg is the mass of the player
V = 8.00 m/s is his velocity
Substituting,
![p'=(110)(8.00)=880 kg m/s](https://tex.z-dn.net/?f=p%27%3D%28110%29%288.00%29%3D880%20kg%20m%2Fs)
The momentum of the ball is
![p=mv](https://tex.z-dn.net/?f=p%3Dmv)
where
m = 0.410 kg is the mass of the ball
v = 25.0 m/s is the velocity of the ball
Substituting,
![p=(0.410)(25.0)=10.3 kg m/s](https://tex.z-dn.net/?f=p%3D%280.410%29%2825.0%29%3D10.3%20kg%20m%2Fs)
The player's momentum is much larger than the ball's momentum. This problem becomes similar to the previous one in the moment when the player catches the ball: at that point, in fact, the velocity of the player-ball system will change such that their total combined momentum will be equal to the total momentum of the two individual objects before the catch.