Question

two pucks are free to slide on a horizontal surface with negligible friction. the figure shows a top view of the pucks, one of mass 4.0 kg and one of mass 3.0 kg, before and after they undergo an elastic collision with each other. before the collision, the 4.0 - kg puck slides at 6.0 m/s in the +x - direction and the 3.0 - kg puck is at rest. after the collision, the 3.0 - kg puck moves with a speed of 2.0 m/s at an unknown angle \\( \theta _ { 3 } \\) measured clockwise from the +x - direction, as indicated, while the 4.0 - kg puck moves at an unknown speed and at an unknown angle \\( \theta _ { 4 } \\) measured counterclockwise from the +x - direction, as indicated. what is the speed of the 4.0 - kg puck after the elastic collision?
a 1.5 m/s
b 4.0 m/s
c 4.5 m/s
d 5.7 m/s

Explanation:

Step1: Apply conservation of kinetic energy

Since the collision is elastic, kinetic energy is conserved. The initial kinetic energy \(K_{i}=\frac{1}{2}m_{4}v_{4i}^{2}\), where \(m_{4} = 4.0\space kg\) and \(v_{4i}=6.0\space m/s\). So \(K_{i}=\frac{1}{2}\times4.0\times6.0^{2}=72\space J\). The final kinetic energy \(K_{f}=\frac{1}{2}m_{4}v_{4f}^{2}+\frac{1}{2}m_{3}v_{3f}^{2}\), with \(m_{3} = 3.0\space kg\) and \(v_{3f}=2.0\space m/s\). Then \(K_{f}=\frac{1}{2}\times4.0\times v_{4f}^{2}+\frac{1}{2}\times3.0\times2.0^{2}=2v_{4f}^{2}+6\).

Step2: Solve for \(v_{4f}\)

Set \(K_{i} = K_{f}\), so \(72=2v_{4f}^{2}+6\). Rearranging gives \(2v_{4f}^{2}=72 - 6=66\), then \(v_{4f}^{2}=33\), \(v_{4f}=\sqrt{33}\approx 5.7\space m/s\).

Answer:

D. \(5.7\space m/s\)