Hockey puck momentum/ collision

[miamirulz29]

Homework Statement

A puck of mass 7 kg moving at 3 m/s strikes an identical puck that is stationery on frictionless ice. After the collision, the first puck leaves with a speed of v1 at an angle of 41.1647 degrees with respect to its original line of motion and the second puck leaves with a speed of v2 at 48.8353 degrees.

Homework Equations

MaVa + MbVb = MaVa' + MbVb'

The Attempt at a Solution

Puck along x-axis:
MVa = MVa' * Cos 41.1647 + MVb' * Cos -48.8353

y-axis:
0 = MVa' * Sin 41.1647 + MVb' * Sin -48.8353


I solved for Va' and got 1.844 (rounded to 3 decimal places), but that answer is wrong. What did I do wrong?

 

[miamirulz29]

Figured it out. Calculator error.

 

[kerimek]

Your approach to solving the problem using the conservation of momentum equation is correct. However, it seems that you have made a small error in your calculation. When solving for Va', you should use the value of Mb (mass of the second puck) instead of Ma. This is because the second puck is initially at rest and therefore has no initial momentum. So the correct equation would be:
MVa = MVa' * cos 41.1647 + 0 * cos -48.8353
Solving for Va', we get a value of 3.427 m/s, which is the correct answer. Keep in mind that when using conservation of momentum, it is important to use the correct masses and velocities for each object in the equation.