A revolutionary war cannon, with a mass of 2010 kg, fires a 21.9 kg ball horizon

Question

A revolutionary war cannon, with a mass of 2010 kg, fires a 21.9 kg ball horizontally. The cannonball has a speed of 106 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired? Answer in units of m/ s =1.5m/s 1)The same explosive charge is used, so the totalenergy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? Answer in units of m/s (the answer is not 0.57m/s ) 2010 kg, fires a 21.9 kg ball horizontally. The cannonball has a speed of 106 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired? Answer in units of m/ s =1.5m/s 1)The same explosive charge is used, so the totalenergy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? Answer in units of m/s (the answer is not 0.57m/s )

Explanation / Answer

2010kg (x) = (21.9 kg)(106m/s)
x = 1.1549m/s

KE = 0.5*m*v^2
KE from cannonball = 0.5*21.9*106^2 = 123034.2 J
KE from cannon = 0.5*2010*1.1549^2 = 1340.4629 J
Total KE =123034.2 +1340.4629 =124374.66 J

All of this energy is now in the canonball.

KE = 0.5*m*v^2
124374.66= 0.5*21.9*v^2
v = 106.57 m/s

Now the question asks how much faster, therefore you subtract the new (106.57m/s) from the original (106m/s) and you are left with 0.575 m/s as your final answer.

So Final answers:

a)1.1549m/s

b)0.575 m/s

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