Answer to Question #261877 in Physics for Charlie

Question #261877

An 8 kg bowling ball is at rest and a 5 kg bowling ball rolls towards it. Upon collision, The heavier ball moves with a velocity of 3 m/s and a lighter ball stops moving. What is the velocity of the lighter ball before collision?


1
Expert's answer
2021-11-07T19:27:38-0500

Given:

"m_1=8\\:\\rm kg"

"m_2=5\\:\\rm kg"

"v_1=0\\:{\\rm m\/s};\\quad v_1'=3\\:\\rm m\/s"

"v_2=?\\:{\\rm m\/s};\\quad v_2'=0\\:\\rm m\/s"


The law of conservation of momentum gives


"m_1v_1-m_2v_2=-m_1v_1'+m_2v_2'"

The velocity of a lighter ball before collision

"v_2=m_1\/m_2v_1'=8\/5*3=4.8\\:\\rm m\/s"


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