In the figure below, a ball of mass m = 50 g is shot with speed vi = 19 m/s into
ID: 3899925 • Letter: I
Question
In the figure below, a ball of mass m = 50 g is shot with speed vi = 19 m/s into a barrel of spring gun of mass M = 242 g initially at rest on a frictionless surface. The ball sticks in the barrel at a point of maximum compression of the spring. Assume that the increase in thermal energy due to friction between the ball and the barrel is negligible.
(a) What is the speed of the spring gun after the ball stops in the barrel?
m/s
(b) What fraction of initial kinetic energy of the ball is stored in the spring?
Here is the image, can't seem to copy it to the question format: http://i.imgur.com/DV1I78h.gif
Explanation / Answer
This can be done neatly using mostly algebra.
Get the final velocity "u" of gun&ball from conservation of momentum;
(m+M)u = mv
u = mv/(m+M)
u=50*19/(292)=3.25m/s
So the kinetic energies before & after are:
Ki = (1/2)mv^2
Kf = (1/2)(m+M)u^2 = (1/2)(m+M)(mv)^2/(m+M)^2 = mKi/(m+M)
The PE is the difference in kinetic energies;
PE = Ki-Kf = Ki - mKi/(m+M) = MKi/(m+M)
But PE is some fraction "f" of initial kinetic energy
PE = fKi
Comparing the two PE eqs gives;
f = M/(m+M)
= 242/292 = .8287