What is the speed of the spring gun after the ball comes to rest

[Destrio]

1. A ball of mass m is projected with speed vi into a barrell of a spring gun of mass M initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. no energy is lost in friction. a) What is the speed of the spring gun after the ball comes to rest in the barrel? b) What fraction of the initial kinetic energy of the ball is lost thorough work done on the spring?

m1v1 = vf(m+M)
Vf = mvi/(M+m)
that's correct

Kfi = (1/2)mvi^2
Kf = (1/2)(m+M)[mvi/(M+m)]^2 = (1/2)m^2vi^2/(m+M)

Kf/Ki = m/(m+M)
the answer says M/(m+M)

is this a mistake, or have I missed something, I don't know how I can get M as a numerator

thanks

 

[Doc Al]

You have to read the questions like a lawyer: They want the fraction that is lost, which is (Kf-Ki)/Ki.

 

[Destrio]

ahhh, i see

thanks very much!

thats the last of my questions

you've been a great help!