A cannonball (1kg) is fired into a cannon (m2 = 4kg) parallel to the ground. Aft
ID: 1608011 • Letter: A
Question
A cannonball (1kg) is fired into a cannon (m2 = 4kg) parallel to the ground. After the cannon catches the ball (inelastic collision), both slide for a short moment across a frictionless floor. They later impact and are slowed to a stop by a spring. Vf = 0 , Vi = 0
Compare the change in momentum of the cannon to that of the ball before and after the catch. Which object has a larger change in momentum or is it the same for both?
I believe it is the same for both?
What is the final velocity of the cannon-and-ball system after the catch (but before the impacting of the spring) if the ball’s initial speed was 20 m/s?
If the spring has a spring constant of 0.5 (k=5), how far does the spring compress after it brings the system to a stop?
Thank you for your help.
Explanation / Answer
here linear momentum remain conserve
m1v1 + m2v2 = (m1+m2)v
initially cannon is at rest v2 = 0
1*v1 = 5v
cannon has larger change in momentum
part b )
v1 = 20 m/s
v = 20/5 = 4 m/s
part c )
using energy conservation
(m1+m2)*v^2/2 = kx^2/2
x = 12.65 meter
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