A small puck with a mass of 0.10 kilograms is sliding in the +x-direction on an air table with an initial speed of 8.0 meters/second. The puck collides with a larger puck that is initially at rest. The larger puck has a mass of 0.40 kilograms. After the collision, the smaller puck has a speed of 3.4 meters/second at an angle of +60° from its original path. The angle made by the larger puck is -35°. What is the x-momentum after the collision if the velocity of the larger puck after the collision is 1.5 meters/second? please help D;

Question

anonymous · Answer

initial x momentum = 0.1kg * 8m/s = 0.8 kg * m/s = final x-momentum

anonymous · Answer

thank you! :)

aum · Answer

If there is friction present, then the momentum will not be conserved.
It looks like the x-momentum after the collision is:
0.1 * 3.4 * cos(60) + 0.4 * 1.5 * cos(35) = ?  Kg-m/s

aum · Answer

The above computation works out to be less than the x-momentum before the collision which was 0.1 * 8 = 0.8 kg-m/s.  Therefore, friction is present here and the momentum is not conserved.

anonymous · Answer

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