Physics Help with elastic collision

[teresalin2004]

Homework Statement

A mass 1.3 kg is initially at rest at the top of a 1.8 meter high ramp. It slides down the frictionless ramp and collides elastically with an unknown mass which is initially at rest. After colliding with the unknown mass, the 1.3 kg mass recoils and achieves a maximum height (altitude) of only 0.4 m going back up the frictionless ramp.
A.
Considering the energy of the 1.3 kg mass just before and just after the elastic collision, how much energy is lost by the 1.3 kg mass?
B.
What is the speed of the 1.3 kg mass just before the elastic collision?
C
What is the speed of the 1.3 kg mass just after the elastic collision?
D
Considering conservation of momentum, what is the momentum of the unknown mass just after the elastic collision?
E
What is the velocity of THE UNKNOWN mass before the elastic collision?
F What is the velocity of the unknown mass just after the elastic collision?
G
What is the mass of the unknown mass in kg?

Homework Equations

a=rw Vr^2
vf=m1v1+m2v2/ m1+m2

The Attempt at a Solution

vf=1.3*.04+1.3+1.8/ 1.2+1.8
1.8*1.3 - 1.2*0.4= 1.82 for Part A
So I know that the steps for B and C will be similar as will E and f but I have no idea how to get started

 

[scottdave]

Homework Equations

a=rw Vr^2

What do you use that formula for? Do you know the formula for Kinetic Energy. They tell you that it slides down a frictionless ramp. This tells you that you don't lose any energy to friction. At the top of the ramp, what is the total energy (potential + kinetic)? How much is Potential Energy? How much is Kinetic?
At bottom of the ramp, what is the total energy? Potential is _____? Kinetic is _____?

Elastic collision means that total Kinetic energy before the collision is equal to total Kinetic energy after.

 

[J Hann]

One useful fact to keep in mind is that in an (linear) elastic collision of two objects is that
the relative speed of approach equals the relative speed of separation after the collision.

 

[neilparker62]

Another method albeit going in the same direction as previous posts:

Energy lost by the falling mass (m1) = energy gained by stationary mass (m2).
Change in momentum (m1) = -(Change in momentum (m2))
2 equations , 2 unknowns: Solve to determine m (m2) and post collision v (m2).

 

[haruspex]

Another method albeit going in the same direction as previous posts:

Energy lost by the falling mass (m1) = energy gained by stationary mass (m2).
Change in momentum (m1) = -(Change in momentum (m2))
2 equations , 2 unknowns: Solve to determine m (m2) and post collision v (m2).

I would say that is exactly the same as scottdave's approach, omitting his preliminary discussion about energy conservation.
J Hann's method is neater. It is a special case of Newton's Experimental Law. It is handy because it avoids the quadratic introduced by the KE equation.

 

[neilparker62]

Hope I'm not missing something but other than determining pre collision and post collision velocity for m1 (√(2gh)) , no quadratic should be needed. The simultaneous equations can be solved by dividing Eqn 1 (KE) by Eqn 2 (momentum) eliminating m2 and leaving us with 1/2 v. Probably that's how we arrive at J. Hann's result in the first place.