A 620-g object traveling at 2.1 m/s collides head-on with a 320-g object traveling in the opposite direction at 3.8 m/s.If the collision is perfectly elastic,what is the change in the kinetic energy of the 620-g object?
A) It loses 0.23 J.
B) It gains 0.69 J.
C) It loses 0.47 J.
D) It loses 1.4 J.
E) It doesn't lose any kinetic energy because the collision is elastic.

Question

Quizplus · Accepted Answer

Explanation:The correct answer is A. The change in kinetic energy of the 620-g object is -0.23 J. This is because the collision is perfectly elastic, which means that the total kinetic energy of the system is conserved. The initial kinetic energy of the system is:$$KE_i = (1/2)mv_1^2 + (1/2)mv_2^2 = (1/2)(0.62 kg)(2.1 m/s)^2 + (1/2)(0.32 kg)(3.8 m/s)^2 = 2.26 J$$The final kinetic energy of the system is:$$KE_f = (1/2)mv_3^2 + (1/2)mv_4^2 = (1/2)(0.62 kg)(v_3)^2 + (1/2)(0.32 kg)(v_4)^2$$where v_3 and v_4 are the velocities of the 620-g and 320-g objects after the collision, respectively.Using the conservation of momentum, we can find that:$$v_3 = (m_1v_1 - m_2v_2)/(m_1 + m_2) = (0.62 kg)(2.1 m/s) - (0.32 kg)(3.8 m/s))/(0.62 kg + 0.32 kg) = 0.98 m/s$$$$v_4 = (m_2v_2 - m_1v_1)/(m_1 + m_2) = (0.32 kg)(3.8 m/s) - (0.62 kg)(2.1 m/s))/(0.62 kg + 0.32 kg) = -2.82 m/s$$Therefore, the change in kinetic energy of the 620-g object is:$$KE_f - KE_i = (1/2)(0.62 kg)(0.98 m/s)^2 + (1/2)(0.32 kg)(-2.82 m/s)^2 - 2.26 J = -0.23 J$$

A 620-G Object Traveling at 2