Conservation laws, let's go!

Classical Mechanics Level 2

Body of mass m 1 m_{1} is moving with some horizontal speed and hits the body of mass m 2 m_{2} which is not moving. What is the ratio of the masses ( m 1 m 2 ) \left(\dfrac{m_{1}}{m_{2}}\right) if the speed of the first body reduced 1,5 times after central elastic collision?


The answer is 5.

Dragan Marković by Dragan Marković , 26, Serbia

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1 solution

P C
Apr 27, 2016

Consider the amount of time Δ t \Delta t between the impact, during this time the internal force overwhelm the external force so the system m 1 − m 2 m_1-m_2 can be called a isolated system. We have this formula for m 1 m_1 after the collision v 1 ′ = ( m 1 − m 2 ) v 1 + m 2 v 2 m 1 + m 2 v'_1=\frac{(m_1-m_2)v_1+m_2v_2}{m_1+m_2} ⇔ v 1 ′ = ( m 1 − m 2 ) v 1 m 1 + m 2 ( s i n c e v 2 = 0 ) \Leftrightarrow v'_1=\frac{(m_1-m_2)v_1}{m_1+m_2} \ (since \ v_2=0) ⇔ 2 3 v 1 = ( m 1 − m 2 ) v 1 m 1 + m 2 \Leftrightarrow \frac{2}{3}v_1=\frac{(m_1-m_2)v_1}{m_1+m_2} ⇔ 1 3 m 1 = 5 3 m 2 \Leftrightarrow \frac{1}{3}m_1=\frac{5}{3}m_2 Therefore m 1 m 2 = 5 \frac{m_1}{m_2}=5

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Very nice solution! I solved it with energy conservation and puls conversation laws, but this way it's quite shorter! :)

Dragan Marković - 5 years, 1 month ago

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