Answer to Question #161854 in Classical Mechanics for Rashinda

Question

A 3kg object moving to the right on a frictionless, horizontal surface with a speed of 2m/s collides head on and sticks to a 2kg object that is initially moving to the left with a speed of 4m/s. After the collision:

A. What is the kinetic energy of the system?

B. What is the momentum of the system?

Answer 1

A) Let's first find the final velocity of the combination of two objects:

$$m_1v_1-m_2v_2=(m_1+m_2)v_f,$$ $$v_f=\dfrac{m_1v_1-m_2v_2}{m_1+m_2},$$ $$v_f=\dfrac{3\ kg\cdot2\ \dfrac{m}{s}-2\ kg\cdot4\ \dfrac{m}{s}}{3\ kg+2\ kg}=-0.4\ \dfrac{m}{s}.$$

The sign minus means that the combination of two objects moves to the left after the collision.

Finally, we can find the kinetic energy of the system after the collision:

$$KE_f=\dfrac{1}{2}mv_f^2=\dfrac{1}{2}\cdot5\ kg\cdot(0.4\ \dfrac{m}{s})^2=0.4\ J.$$

B) We can find the momentum of the system after the collision as follows:

$$p_f=(m_1+m_2)v_f,$$ $$p_f=(3\ kg+2\ kg)\cdot(-0.4\ \dfrac{m}{s})=-2.0\ \dfrac{kg\cdot m}{s}.$$