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)The energy of the object is unchanged.

Question

Quizplus · Accepted Answer

Explanation:The total kinetic energy before the collision is:$$KE_i = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{1}{2}(0.620 kg)(2.1 m/s)^2 + \frac{1}{2}(0.320 kg)(3.8 m/s)^2 = 2.72 J$$The total kinetic energy after the collision is:$$KE_f = \frac{1}{2}mv_3^2 + \frac{1}{2}mv_4^2$$where v3 and v4 are the velocities of the two objects after the collision. Since the collision is perfectly elastic, the total kinetic energy is conserved, so:$$KE_i = KE_f$$$$2.72 J = \frac{1}{2}(0.620 kg)v_3^2 + \frac{1}{2}(0.320 kg)v_4^2$$Solving for v3, we get:$$v_3 = \pm 1.4 m/s$$Since the 620-g object is traveling in the opposite direction after the collision, its velocity is -1.4 m/s. Therefore, the change in its kinetic energy is:$$\Delta KE = KE_f - KE_i = \frac{1}{2}(0.620 kg)(-1.4 m/s)^2 - 2.72 J = -0.23 J$$So, the correct answer is A: It loses 0.23 J.

A 620-G Object Traveling at 2