Answer to Question #108569 in Electricity and Magnetism for juju

Question #108569
A 0.015kg marble moving to the right at 0.225m/s makes an elastic head-on collision with a 0.030kg at 0.180m/s. After a collision, the smaller marble moves to the left at 0.315m/s. assume that neither marble rotates before or after the collision and both marbles are moving on a frictionless surface. What is the velocity of the 0.030kg marble after the collision?
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Expert's answer
2020-04-08T10:42:30-0400

Solution: We will use the law of momentum conservation. The initial momentum of first (small) marble is P1=m1v1\vec P_1=m_1\cdot \vec v_1 . That is for second marble P2=m2v2.\vec P_2=m_2\cdot \vec v_2 . After elastic collision the momentum of smaller marble is P1new=m1V1\vec P_{1new}=m_1\cdot \vec V_1 and greater one P2new=m2V2\vec P_{2new}=m_2\cdot \vec V_2 . Due to the fact that the collision is elastic head-on and without rotation, all vectors are aligned relative to the same direction and differ only in the sign. The law of momentum conservation state

(1)P1+P2=P1new+P2new(1) P_1+P_2=P_{1new}+P_{2new} substituting formulas for momentums we have

(2) m1v1m2v2=m1V1+m2V2m_1\cdot v_1- m_2\cdot v_2=-m_1\cdot V_1+m_2\cdot V_2 . Here we have taken into account the + sign for right moving and the - sign for left moving . We solve this equation with respect to an unknown speed V2V_2

V2=m1v1m2v2+m1V1m2=m1m2(v1+V1)v2==0.015kg0.030kg(0.225+0.315)ms10.18ms1=0.09m/sV_2=\frac{m_1\cdot v_1- m_2\cdot v_2+m_1\cdot V_1}{m_2}=\frac{m_1}{m_2}(v_1+V_1)-v_2=\\=\frac{0.015kg}{0.030kg}(0.225+0.315)ms^{-1}- 0.18ms^{-1}=0.09m/s

Answer: The velocity of the 0.030kg marble after the collision is 0.09m/s to the right.


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