A 2180 kg cannon fires a 17.3 kg ball horizontally?

Question

The cannonball has a speed of 133 m/s after it left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally.
What is the speed of the cannon just after it was fired in units of m/s?

Assume the same explosive charge is used and the total energy of the cannon plus the cannonball system remains the same. Disregard friction. How much FASTER would the ball travel if the cannon was mounted rigidly but all other parameters were the same (in units of m/s)?

Answer 1

The first problem deals with conservation of momentum. Momentum is equal to mass times velocity.
m(1)*v(1) = m(2)*v(2)

m(1) = 2180 kg
v (1) = ???
m(2) = 17.3 kg
v (2) = 133 m/s

2180*x = 17.3*133
x = 1.06 m/s

The second problem deals with conservation of energy. Basically, the problem states that all the energy is conserved (which never happens in real life). All the energy in this problem is kinetic energy.

KE = 0.5*m*v^2
KE from cannonball = 0.5*17.3*133^2 = 153,010 J
KE from cannon = 0.5*2180*1.06^2 = 1,214 J
Total KE = 153,010 + 1,214 = 154,224 J

All of this energy is now in the canonball.

KE = 0.5*m*v^2
154,224 = 0.5*17.3*v^2
v = 133.5 m/s

Answer 2

Part 2 is is also a momentum, not energy problem. By completing part 1, you already have the answer:

Assuming the 133 m/s is relative to a stationary coordinate system and the canon's 1.06 m/s is the opposite direction in the same system, the new speed of the ball is clearly just the sum of the 2 speeds--- 134.06 m/s

QED

Answer 3

I agree with the solution of Josh, except that he did not answer the second part of the problem. It doesn't ask for the speed of the cannonball, but how much FASTER would it travel. So the answer is 0.5m/s calculated as 133.5-133.

Answer 4

500 m/s.