Answer to Question #179652 in Physics for Robert

Question #179652

Two balls weighing 2.5 kg and 5 kg are subject to an elastic collision. Before the collision, the first and lighter ball was at rest, and the second one was moving at a speed of 3.5 m/s. Determine the kinetic energy of a lighter ball after a collision. (Answer: V1'=3.66 m/s, V2' = 1.166 m/s and Ek1'= 16.7 J)


1
Expert's answer
2021-04-19T07:22:00-0400

Let v1=0,v2=3.5m/sv_1 = 0, v_2 = 3.5m/s be the speeds of the balls before the collision, m1=2.5kg,m2=2m1=5kgm_1 = 2.5kg, m_2 = 2m_1 = 5kg the masses of the balls, v1,v2v_1', v_2' the speeds of the balls after the collision.

Suppose the collision was head-on and ellastic. Thus, according to the momentum conservation law, have:


m1v1+m2v2=m1v1+m2v2m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'

Since v1=0v_1 = 0:


m2v2=m1v1+m2v22m1v2=m1v1+2m1v22v2=v1+2v2m_2v_2 = m_1v_1' + m_2v_2'\\ 2m_1v_2 = m_1v_1' + 2m_1v_2'\\ 2v_2 = v_1' + 2v_2'

According to the energy conservation law:


m2v222=m1v122+m2v2222m1v222=m1v122+2m1v2222v22=v12+2v22\dfrac{m_2v_2^2}{2} = \dfrac{m_1v_1'^2}{2} + \dfrac{m_2v_2'^2}{2}\\ \dfrac{2m_1v_2^2}{2} = \dfrac{m_1v_1'^2}{2} + \dfrac{2m_1v_2'^2}{2}\\ 2v_2^2 = v_1'^2 + 2v_2'^2

Expressing v1v_1' from 2v2=v1+2v22v_2 = v_1' + 2v_2' and substituting it into 2v22=v12+2v222v_2^2 = v_1'^2 + 2v_2'^2, obtain:


3v224v2v2+v22=03v_2'^2 - 4v_2v_2' + v_2^2 = 0

Substituting the values and solving the quadratic equation, obtain:


v2=1.166m/s or v2=3.5m/sv_2' = 1.166 m/s\space \text{or}\space v_2' = 3.5 m/s

The last values in impossible, since it would mean that the first ball remains at rest after the collsion. Thus, subsituting v2=1.166m/sv_2' = 1.166 m/s into 2v2=v1+2v22v_2 = v_1' + 2v_2', obtain:


v1=2(v2v2)=4.66m/sv_1' = 2(v_2 - v_2') = 4.66m/s

And the kinetic energy is:


E1=m1v122=27.2JE_1' = \dfrac{m_1v_1'^2}{2} = 27.2J

Answer. 27.2J


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