A 1.9kg block slides along a frictionless surface at 2.0 m/s. A second block, sl
ID: 1695264 • Letter: A
Question
A 1.9kg block slides along a frictionless surface at 2.0 m/s. A second block, sliding at a faster 4.7 m/s, collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/s.What was the mass of the second block?
-I know the equation to use is m1*v1 + m2*v2 = (m1+m2)* vf, but I cant figure out how to solve this equation for m2. If someone could please show me how to do it step by step that would be amazing!!! Or tell me if I should be using something else too.
Explanation / Answer
before collision: mass m_1 =1.9 kg speed u_1 =2 m/s mass m_2 =? kg speed u_2 =4.7 m/s after collision: speed v_f =2.5 m/s apply conservation of momentum, m_1u_1+m_2u_2 =(m_1v_f+m_2v_f) m_1u_1-m_1v_f =m_2v_f-m_2u_2 m_1(u_1-v_f) =m_2(v_f-u_2) m_2 =[m_1(u_1-v_f)] /[(v_f-u_2)] substitute the given data in above equation we get m_2 =0.43 kg
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