A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest.If after the collision the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, what is the final velocity of the first puck?
A)1.0 m/s, 30° below the x axis
B)2.3 m/s, 50° below the x axis
C)2.8 m/s, 60° below the x axis
D)3.0 m/s, 45° below the x axis
E)not enough information

Question

Quizplus · Accepted Answer

Since momentum is conserved in the collision, we can write:
$m v_{1i} = m v_{1f} \cos	heta_1 + m v_{2f} \cos	heta_2$
where $v_{1i}$ is the initial velocity of the first puck, $v_{1f}$ is its final velocity, $v_{2f}$ is the final velocity of the second puck, and $	heta_1$ and $	heta_2$ are the angles the velocities make with the x-axis. 
Also, we know that the total kinetic energy before and after the collision must be the same, so we can write:
$\frac{1}{2} m v_{1i}^2 = \frac{1}{2} m v_{1f}^2 + \frac{1}{2} m v_{2f}^2$
Now, we can substitute the given values and solve for $v_{1f}$:
$m \cdot 4.5 	ext{ m/s} = m v_{1f} \cos 0^\circ + m \cdot 3.5 	ext{ m/s} \cos 30^\circ$
$\frac{1}{2} m \cdot (4.5 	ext{ m/s})^2 = \frac{1}{2} m v_{1f}^2 + \frac{1}{2} m \cdot (3.5 	ext{ m/s})^2$
Simplifying, we get:
$v_{1f} = 2.3 	ext{ m/s}$, 50° below the x-axis. So, the answer is (B).

A Hockey Puck of Mass M Traveling Along the X