The recoil speed of the cannon is 9 m/s
Explanation:
We can solve the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon+cannonball system before and after the shot must be conserved, in absence of external forces.
Before the shot, the total momentum is zero:
p = 0
After the shot, the total momentum is: given by:
[tex]p=mv+MV[/tex]
where :
m = 10 kg is the mass of the ball
v = 90 m/s is the velocity of the ball
M = 100 kg is the mass of the cannon
V is the recoil velocity of the cannon
Since momentum is conserved, we can write
[tex]0=p = mv+MV[/tex]
And solving for V, we find the recoil velocity of the cannon:
[tex]0=mv+MV\\V=-\frac{mv}{M}=-\frac{(10)(90)}{100}=-9 m/s[/tex]
And the negative sign means the cannon will move in the opposite direction to the cannonball: so, the recoil speed is 9 m/s.
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