Elastic collision between two moving objects

[captainmustard]

Homework Statement

A 0.2 kg block, moving at 6 m/s, is catching up and colliding elastically with a 0.6 kg block that is moving along the same line and in the same direction. Find the velocities of the ball after this one-dimensional collision.

Homework Equations

Conservation of momentum, conservation of energy

The Attempt at a Solution

First, I figured the momentum and kinetic energy would be conserved so I setup a couple of equations:

m1v1i + m2v2i = m1v1f + m2v2f
and
m1v1i^2 + m2v2i^2 = m1v1f^2 + m2v2f^2

At this point, I realize I have 3 unknowns (v2i, v1f and v2f) and I simply don't know what I can even do with this information. Any ideas at all would be helpful.

 

[TSny]

Hi, captainmustard. Welcome to PF!

You are right. There is not enough information to answer the question. You have set up the correct equations.

 

[captainmustard]

Thank you for the response. I thought I was going crazy for a moment there. Would this be possible if it were a completely inelastic collision? I would imagine it would involve some tedious algebra if so.

 

[ehild]

No, you need the initial velocity of the 0.6 kg ball, too.

ehild

 

[Nickie]

I would approach this problem by first defining the variables and their units. The mass of the first block, m1, is 0.2 kg and its initial velocity, v1i, is 6 m/s. The mass of the second block, m2, is 0.6 kg and its initial velocity, v2i, is also 6 m/s. The final velocities of both blocks after the collision are v1f and v2f.

Next, I would use the equations of conservation of momentum and conservation of energy to solve for the unknowns. In an elastic collision, both momentum and kinetic energy are conserved.

Using the conservation of momentum equation, we can write:

m1v1i + m2v2i = m1v1f + m2v2f

Substituting in the known values, we get:

0.2(6) + 0.6(6) = 0.2(v1f) + 0.6(v2f)

Simplifying, we get:

1.2 = 0.2(v1f) + 0.6(v2f)

Next, using the conservation of energy equation, we can write:

m1v1i^2 + m2v2i^2 = m1v1f^2 + m2v2f^2

Substituting in the known values, we get:

0.2(6)^2 + 0.6(6)^2 = 0.2(v1f)^2 + 0.6(v2f)^2

Simplifying, we get:

7.2 = 0.2(v1f)^2 + 0.6(v2f)^2

Now we have two equations and two unknowns (v1f and v2f). We can solve for these unknowns by using algebraic manipulation or by using a graphing calculator.

Solving for v1f in the first equation, we get:

v1f = (1.2 - 0.6(v2f))/0.2

Substituting this into the second equation, we get:

7.2 = 0.2[(1.2 - 0.6(v2f))/0.2]^2 + 0.6(v2f)^2

Simplifying, we get a quadratic equation:

(v