1) -0.5 m/s
We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.
Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:
[tex]p=0[/tex]
After the shot, the momentum is:
[tex]p=MV+mv[/tex]
where
M = 2000 kg is the mass of the cannon
m = 10 kg is the mass of the shell
v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)
V = ? is the velocity of the cannon
Since momentum is conserved, we can write
[tex]0=MV+mv[/tex]
And solving for V, we find the velocity of the cannon:
[tex]V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s[/tex]
where the negative sign indicates that the cannon moves in the direction opposite to the shell.
2) 0.5 m
The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity of the cannon
u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)
[tex]a=-0.25 m/s^2[/tex] is the deceleration of the cannon
s is the distance travelled by the cannon
The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m[/tex]
