Question

A hockey puck of mass m traveling along the x axis at 6.0 m/s hits another identical hockey puck at rest. If after the collision the second puck travels at a speed of 4.8 m/s at an angle of 30° above the x axis, what is the final velocity of the first puck

Answer 1

Answer:

Velocity is 3.02 m/s at an angle of 53.13° below X-axis.

Explanation:

Let unknown velocity be v.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 6i + m x 0i = 6m i

Final momentum = m x (4.8cos 30 i + 4.8sin 30 j) + m x v = 4.16 m i + 2.4 m j + m v

Comparing

4.16 m i + 2.4 m j + m v = 6m i

v = 1.84 i - 2.4 j

Magnitude of velocity

       $v=\sqrt{1.84^2+(-2.4)^2}=3.02m/s$

Direction,  

        $\theta =tan^{-1}\left ( \frac{-2.4}{1.8}\right )=-53.13^0$     

Velocity is 3.02 m/s at an angle of 53.13° below X-axis.