# Momentum question involving an elastic collision

1. A 62 kg skater traveling at 2.8 m/s accidentally bumps elastically into a 76 kg skater moving at 1.6m/s in the same direction
Find the velocity of each skater after the collision

2. m1v1 + m1v2 = m1v3 + m2v4

3.

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As you stated, momentum is conserved but so is one other thing. Can you figure out what?

Kinetic energy is also conserved.

This is all the work I have done so far.

(62)(2.8) + (76)(1.6) = 62v1 + 76v2

0.5(62)(2.8)^2 + 0.5(76)(1.6)^2 = 0.5(62)(v1)^2 + 0.5(76)(v2)^2

295.2 = 62v1 + 76v2 and 340.32 = (31)(v1)^2 + (38)(v2)^2

v2= (295.2 - 62v1)/76

340.32 = (31)(v1)^2 + (38)[(295.2 - 62v1)/76]^2

0 = (31)(v1)^2 + 357.778v1 + (25.289)(v1)^2 + 573.30947 - 340.32

Last edited:
BruceW
Homework Helper
It all looks good except the last line. In fact, just the term with v1 looks incorrect. Try checking the bracket expansion again?

I fixed it.
When I solve it I get -0.7366 and -5.6195 for v1 but neither of those are answers.

NascentOxygen
Staff Emeritus
I fixed it.
When I solve it I get -0.7366 and -5.6195 for v1 but neither of those are answers.

There is always the option of using an online calculator to check your answer, to give yourself confidence that you have the solution, e.g., http://www.quickmath.com/webMathema...+(31)(v1)^2+++(38)((295.2+-+62v1)/76)^2&v2=v1

Note: Mathematica ignores square brackets, so you must change them to round parentheses ().

ehild
Homework Helper
If you set up the equation correctly, one root has to be the same as the original velocity.

You can follow a much simpler method of solving.

Arrange the equations both for momentum and energy so that the initial and final velocity of one body is at one side and those of the other body on the other side:

m1(u1-v1)=m2(v2-u2)
m1(u12-v12)=m2(v22-u22).

As v≠u for either body, you can divide the second equation with the first one, and obtain a simple first-order system of equations

u1+v1=u2+v2,
m1(u1-v1)=m2(v2-u2).

ehild