Question

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?

Answer 1

Solution: We will use the law of momentum conservation. The initial momentum of first (small) marble is $\vec P_1=m_1\cdot \vec v_1$ . That is for second marble $\vec P_2=m_2\cdot \vec v_2 .$ After elastic collision the momentum of smaller marble is $\vec P_{1new}=m_1\cdot \vec V_1$ and greater one $\vec P_{2new}=m_2\cdot \vec V_2$ . 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

$(1) P_1+P_2=P_{1new}+P_{2new}$ substituting formulas for momentums we have

(2) $m_1\cdot v_1- m_2\cdot v_2=-m_1\cdot V_1+m_2\cdot V_2$ . 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 $V_2$

$V_2=\frac{m_1\cdot v_1- m_2\cdot v_2+m_1\cdot V_1}{m_2}=\frac{m_1}{m_2}(v_1+V_1)-v_2=\\=\frac{0.015kg}{0.030kg}(0.225+0.315)ms^{-1}- 0.18ms^{-1}=0.09m/s$

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