# Homework Help: The velocity of each object after the collision

1. Jul 11, 2010

### mizzy

1. The problem statement, all variables and given/known data
A 5g object moving to the right with a speed of 20cm/s makes an elastic head on collision with a 10g object initally at rest. Find a) The velocity of each objet after the collision. b) The fraction of the energy transferred to the 10g object.

2. Relevant equations
In an elastic collision, momentum and energy is conserved.

m1v1 + m2v2 (before collision) = m1v1 + m2v2 (after collision)

v1i - v2i = -(v1f - v2f)

3. The attempt at a solution
where do i start? I tried using the conservation of momentum equation, but you end up with two unknowns, final velocities of both objects.

2. Jul 11, 2010

### Staff: Mentor

But you have two equations. Use them both to solve for the final velocities. (The second equation that you list incorporates energy conservation, valid for elastic collisions.)

3. Jul 11, 2010

### mizzy

ok, i have my two equations.

first equation :
0.001 = (0.005)V1f + (0.01)V2f

second equation:
0.2 = -V1f + V2f

i might need a refresher is solving simultaneous equations. Can I multiply 0.01 to the second equation?

4. Jul 11, 2010

### Staff: Mentor

Good.
Sure. Then you'd subtract the two equations to eliminate V2f. (You can also multiply the second equation by 0.005 and add the two equations to eliminate V1f.)

5. Jul 11, 2010

### mizzy

Thank You!

6. Jul 11, 2010

### mizzy

for part b, when it asks for the fraction of energy transferred to the 10g object, is it just the kinetic energy before the collision?

7. Jul 11, 2010

### Staff: Mentor

Compare the initial KE before collision to the KE of the 10g object after collision. What fraction of that initial KE went into the 10g object?

8. Jul 11, 2010

### mizzy

Oh ok. So what I did was to find the kinetic energy before the collision and also the kinetic energy of the 10g object. Then I took the kinetic energy of the 10g object and divided it by the kinetic energy before the collision.

is that correct?

9. Jul 12, 2010

### Staff: Mentor

Yes, that's correct.