A hockey puck with a mass of 160 g slides at a speed of 12 m/s on frictionless ice.The puck runs into a small box with a mass of 850 g that is initially at rest.Assuming that the collision is elastic,what is the speed of the box after the collision?
A)2.3 m/s
B)3.8 m/s
C)5.6 m/s
D)6.0 m/s
E)8.2 m/s

Question

Quizplus · Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. The formula for conservation of momentum is $m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$, where $m_1$ and $m_2$ are the masses of the two objects, $v_1$ and $v_2$ are their initial velocities, and $v_1'$ and $v_2'$ are their final velocities. Since the box is initially at rest, $v_2 = 0$. We're solving for $v_2'$, the final velocity of the box.Given:- $m_1 = 160g = 0.16kg$- $m_2 = 850g = 0.85kg$- $v_1 = 12m/s$- $v_2 = 0m/s$ (initially at rest)Using the conservation of momentum and knowing that in an elastic collision, the relative speed of approach before collision is equal to the relative speed of separation after collision, we can derive the formula for the final velocity of the second object (the box) in an elastic collision:$$v_2' = \frac{2m_1v_1 + v_2(m_2 - m_1)}{m_1 + m_2}$$Substituting the given values:$$v_2' = \frac{2(0.16kg)(12m/s) + (0m/s)(0.85kg - 0.16kg)}{0.16kg + 0.85kg}$$$$v_2' = \frac{3.84}{1.01}$$$$v_2' \approx 3.8 m/s$$Therefore, the speed of the box after the collision is approximately 3.8 m/s.

A Hockey Puck with a Mass of 160 G Slides