A) -12 m/s
B) -360 N
Explanation:
A)
We can solve this problem by using the law of conservation of momentum: in fact, the total momentum of the plunger + bullet system must be conserved.
Initially, their total momentum is zero, since they are at rest:
[tex]p_i=0[/tex]
While the final total momentum is
[tex]p_f=mv+MV[/tex]
where:
m = 0.02 kg is the mass of the bullet
v = 600 m/s is the velocity of the bullet
M = 1 kg is the mass of the plunger
V is the recoil velocity of the plunger
Since momentum is conserved,
[tex]p_i=p_f[/tex]
And so we find V:
[tex]0=mv+MV\\V=-\frac{mv}{M}=-\frac{(0.02)(600)}{1}=-12 m/s[/tex]
B)
From part A), we said that the speed of the plunger after the shot is
u = 12 m/s
In order to be stopped, its final velocity must be
v = 0
Since its an accelerated motion, we can find its acceleration using the suvat equation
[tex]v^2-u^2=2as[/tex]
where
s = 20 cm = 0.20 m is the stopping distance
Solving for the acceleration, we find
[tex]a=\frac{v^2-u^2}{2s}=\frac{0-12^2}{2(0.20)}=-360 m/s^2[/tex]
And since the mass of the plunger is
m = 1 kg
The force on it would be (Newton's second law)
[tex]F=ma=(1)(-360)=-360 N[/tex]
