# Elastic collision with a spring constant and unknown masses

## Homework Statement

A ball of mass m rolls down a 3.0 m ramp inclined at 30° above the horizontal, rolls along a flat, friction less surface, and collides elastically with another ball of mass 2m, initially at rest. The second mass then moves along the surface and collides with a horizontally mounted spring with spring constant k= 200 N/m, compressing it by 0.15 m. What is the value of m?

## Homework Equations

Ek 1 + Eg 1 + Ee 1 = Ek 2 + Eg 1 + Ee 2

For the incline:
FN + Fg parallel + Fg perpendicular + Ff = m1a

## The Attempt at a Solution

I've been working backwards so I started with this:

1/2 m2v2' 2 = 1/2 kx2

I ended up with v2' = √45/2m and I now have no idea how to continue

Any help will be appreciated.

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
The top of the 3 m long, 30 degree ramp is 3 sin(30)= 1.5 m above its base. Taking the kinetic energy of the first ball to be 0 at the top of the ramp, its potential energy is 1.5mg, relative to the bottom. At the bottom of the ramp, it potential energy is 0 so its kinetic energy is 1.5mg and its velocity is given by $v= \sqrt{1.5g}$ so its momentum is $m\sqrt{1.5g}$. On the level, both "conservation or kinetic energy" and "conversation of momentum" hold.

Oops. I forgot about conservation of momentum. Thanks!