# Homework Help: Two block collision (momentum)

1. Nov 7, 2012

### lemin_rew

1. The problem statement, all variables and given/known data

a)Two blocks are moving on a frictionless surface. The first block's mass is 1.07kg and its initial velocity is 5.98m/s. The second block's mass is 2.26kg and its initial velocity is 2.59m/s. What is the velocity of the 1.07kg block after the collision if the velocity of the second block is 5.15m/s after the collision?

b)If the initial velocity of the 2.26kg block is the same magnitude but in the opposite direction, what is the velocity of the 1.07kg block after the collision?
2. Relevant equations

m1v1=m2v2

3. The attempt at a solution

a)m1(vf+vi)=m2(vf+vi)
i made the velocities positive, because they travel n the same direction even after the collision.

(1.07kg)(vf+5.98m/s)=(2.26kg)(2.59m/s+5.15m/s)
vf= 10.37m/s

b) i will try after i get the first part right

2. Nov 7, 2012

### LawrenceC

The correct formula for conservation of momentum is:

m1*V1i + m2*V2i = m1*V1f + m2*V2f where
i is initial velocity before collision
f is final velocity after collision

It is not what you have written.

3. Nov 7, 2012

### lemin_rew

oh okay!! thank you so much.
then would the second part be the same equation but with the opposite direction.
so, m1*V1i + m2*-V2i = m1*V1f + m2*-V2f
like this?

4. Nov 7, 2012

### LawrenceC

For the second part, b, no final velocity is given as was the case for part a. In situation b, the momentum equation has two unknowns, namely, the final velocity of each block. You need another equation.

Is the collision perfectly elastic?

5. Nov 7, 2012

### lemin_rew

would it be then:
V1f=((m1-m2)/(m1+m2))v1i + ((2m2)/(m1+m2))/v2i

6. Nov 7, 2012

### LawrenceC

For part b the blocks are heading towards one another initially. Therefore for conservation of momentum you have the equation:

m1*V1i + m2*V2i = m1*V1f + m2*V2f

where V1i and V2i have opposite signs. However, you cannot solve the equation because you have two unknowns, namely V1f and V2f. You need another equation with the same two unknowns. That equation comes from conservation of energy providing the collision is perfectly elastic. Kinetic energy is conserved in an elastic collision.

7. Nov 7, 2012

### lemin_rew

so, then kinetic energy plays a role and the equation is:
1/2m1v1i^2 + 1/2m2v2i = 1/2m1v1f^2 + 1/2m2v2f^2
where v1i and v2i have opposite signs?

8. Nov 7, 2012

### LawrenceC

For the kinetic energy, the velocity is squared so the sign does not matter. Kinetic energy is always positive.

1/2m1v1i^2 + 1/2mv2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2

So now you have two equations and two unknowns. Substitute for one of the final velocities from the momentum equation which leaves you with a quadratic to solve.

9. Nov 7, 2012

### lemin_rew

oh i think i get what you mean, let me try
thank you

Last edited: Nov 7, 2012
10. Nov 7, 2012

### lemin_rew

i got my parabola to be
22.18v1f^2-0.257v1f-26.6463
and my root is 1.10.

11. Nov 8, 2012

### LawrenceC

I do not get the same numerical answer as you. Plug your numerical results back into both the momentum and energy equations to see if you did the arithmetic correctly.

12. Nov 8, 2012

### LawrenceC

I should add that of the two solutions you get from the quadratic equation, one should be the initial velocity before the collision; the other should be the velocity after the collision. When you plug your quadratic solutions back into the momentum equation, you should get the corresponding velocities of the other block before and after the collision.