A ball with a mass of # 2 kg# is rolling at #2 m/s# and elastically collides with a resting ball with a mass of #9 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of # 2   kg# is rolling at #2   m/s# and elastically collides with a resting ball with a mass of #9  kg#. What are the post-collision velocities of the balls?

Julia Aguilar · Accepted Answer

The initially moving ball would move #1.273 m/s# to the left while the ball initially at rest would move #0.72m/s# to the right.

(NOTE: The final velocity of an arbitrary object is indicated by velocities with apostrophes.) Examine the system's overall momentum. #mv_1+mv_2 = mv'_1+mv'_2 # Note that final velocities are positive because I assumed them to move to the right after the impact. #mv_1+mcancel(v_2) = mv'_1+mv'_2 # #(2)(2) = (2)(v'_1)+(9)(v'_2 )# Moreover, the problem states that the impact involves an elastic collision. Therefore, the coefficient of restitution of the system, #e#, is 1. #e = 1=(v'_2-v'_1)/(v_1-v_2) = (v'_2-v'_1)/(2-0) # #therefore,2 =v'_2-v'_1# To solve the post-velocities solve the system of equations: #(1) : 9v'_2 +2v'_1=4# #(2) : v'_2-v'_1=2# Where #v'_1= -1.273# and #v'_2 = +0.72# Since #v'_1# is negative, the assumed direction of its final velocity is the opposite.

William Audy · Answer

undefined To find the post-collision velocities of the balls, we can use the conservation of momentum and the conservation of kinetic energy.

Let $ v_1 $ be the velocity of the 2 kg ball after the collision and $ v_2 $ be the velocity of the 9 kg ball after the collision.

From the conservation of momentum:

$$ m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2 $$

where $ m_1 = 2 $ kg, $ m_2 = 9 $ kg, $ u_1 = 2 $ m/s, and $ u_2 = 0 $ m/s.

From the conservation of kinetic energy:

$$ \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 $$

Solving these equations simultaneously, we can find $ v_1 $ and $ v_2 $.

After solving, we get $ v_1 = 1.1 $ m/s and $ v_2 = 0.33 $ m/s.