Answer to Question #143596 in Physics for mj

Question

A 0.015-kg marble moving to the right at 0.40 m/s has a head-on, elastic collision with a 0.045 kg marble sitting at rest on a smooth level surface. Which of the following are the correct magnitudes and directions of the velocities of the two marbles after collision?

Answer 1

From the Law of Conservation of Momentum, we have:

$$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f},$$

here, $m_1=0.015\ kg$ is the mass of the first marble, $m_2=0.045\ kg$ is the mass of the second marble, $v_{1i}=0.40\ \dfrac{m}{s}$ is the initial velocity of the first marble before the collision, $v_{2i}=0\ \dfrac{m}{s}$ is the initial velocity of the second marble before the collision, $v_{1f}$ is the final velocity of the first marble after the collision, $v_{2f}$ is the final velocity of the second marble after the collision.

Since collision is elastic, kinetic energy is conserved and we can write:

$$\dfrac{1}{2}m_1v_{1i}^2+\dfrac{1}{2}m_2v_{2i}^2=\dfrac{1}{2}m_1v_{1f}^2+\dfrac{1}{2}m_2v_{2f}^2.$$

This formula gives us an additional relationship between velocities. Therefore, with the help of these two formulas we can find the velocity of each marbles after collision:

$$v_{1f}=\dfrac{(m_1-m_2)}{(m_1+m2)}v_1,$$ $$v_{1f}=\dfrac{(0.015\ kg-0.045\ kg)}{(0.015\ kg+0.045\ kg)}\cdot 0.40\ \dfrac{m}{s}=-0.20\ \dfrac{m}{s}.$$

The sign minus means that the first marble moves to the left at $0.20\ \dfrac{m}{s}.$

$$v_{2f}=\dfrac{2m_1v_1}{(m_1+m_2)},$$ $$v_{2f}=\dfrac{2\cdot 0.015\ kg\cdot 0.40\ \dfrac{m}{s}}{(0.015\ kg+0.045\ kg)}=0.20\ \dfrac{m}{s}.$$

The sign plus means that the second marble moves to the right at $0.20\ \dfrac{m}{s}.$

Answer:

$v_{1f}=0.20\ \dfrac{m}{s},$ left.

$v_{2f}=0.20\ \dfrac{m}{s},$ right.