Total momentum after collision

[_mae]

Excuse the Title =X

Homework Statement

A 0.300 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200 kg puck moving initially along the x-axis with a speed of 2.00 m/s. After the collision, the 0.200 kg puck has a speed of 1.00 m/s at an angle of $$\Theta$$= 50.0° to the positive x axis.

http://img265.imageshack.us/img265/5695/811vx2.gif

Determine the speed of the 0.300 kg puck after the collision.​

Homework Equations

m1v1i + m2v2i = (m1 + m2)vf

The Attempt at a Solution

total momentum before collision = 0
total momentum after collision = 0.3 * v1f *sin 50 - 0.2 * v2f *sin φ
from the law of conservation of momentum ,
0.3 * v1f * sin 50 - 0.2 * v2f * sin φ = 0
0.3 * v1f * sin 50 = 0.2 * v2f * sin φ
where v2f = 1 m / s
φ = 90 - 50 = 40

so: (0.3)(v1f)(sin50) - (0.2)(1)(sin40) = 0.5596 m/s ??

 

[rl.bhat]

total momentum before collision = 0 This statement is wrong, because 0.2 kg mass is moving with a velocity 2 m/s.
total momentum after collision = 0.3 * v1f *sin 50 - 0.2 * v2f *sin φ This is only y-components. What about x-components?

 

[ogb p]

Your calculation is correct. The total momentum before and after the collision must be equal according to the law of conservation of momentum. Therefore, the speed of the 0.300 kg puck after the collision is 0.5596 m/s. Great job!