(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:
[tex]p=mv+MV[/tex] (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:
[tex]0=mv+MV\\V=-\frac{mv}{M}=-\frac{(0.0250)(550)}{3.00}=-4.6 m/s[/tex]
And the negative sign means the rifle will move backward.
(b) 31.7 J
The kinetic energy of an object is given by
[tex]K=\frac{1}{2}mv^2[/tex]
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
[tex]K=\frac{1}{2}(3.00)(4.6)^2=31.7 J[/tex]
(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:
[tex]0=mv+MV[/tex]
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:
[tex]V=-\frac{mv}{M}=-\frac{(0.0250)(550)}{28.0}=-0.5 m/s[/tex]
(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
[tex]K=\frac{1}{2}mv^2[/tex]
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,
[tex]K=\frac{1}{2}(28.0)(0.5)^2=3.5 J[/tex]
(e) Player's momentum is larger
The momentum of the player is
[tex]p'=MV[/tex]
where
M = 110 kg is the mass of the player
V = 8.00 m/s is his velocity
Substituting,
[tex]p'=(110)(8.00)=880 kg m/s[/tex]
The momentum of the ball is
[tex]p=mv[/tex]
where
m = 0.410 kg is the mass of the ball
v = 25.0 m/s is the velocity of the ball
Substituting,
[tex]p=(0.410)(25.0)=10.3 kg m/s[/tex]
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.
