A revolutionary war cannon, with a mass of 2240 kg, fires a 22.6 kg ball horizon

Question

A revolutionary war cannon, with a mass of 2240 kg, fires a 22.6 kg ball horizontally. The cannonball has a speed of 120 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.

The same explosive charge is used, so the total energy 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 re- mained the same?

Answer in units of m/s.

Many thanks

Explanation / Answer

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) = 2240kg
v (1) = ???
m(2) = 22.6 kg
v (2) = 120m/s

2240*x = 22.6*120
x = 1.21m/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*22.6*120^2 = 162720 J
KE from cannon = 0.5*2240*1.21^2 =1639.792 J
Total KE = 162720+ 1639.792 = 164359.792 J

All of this energy is now in the canonball.

KE = 0.5*m*v^2
164359.792 = 0.5*22.6*v^2
v =120.603 m/s

extra speed = 0.603m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.