Answer to Question #261881 in General Chemistry for Charlie

Question

An 8 kg bowling ball is at rest and a 5 kg bowling ball rolls toward 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?

Answer 1

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

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