Question

before after? θ₄ +y 6.0 m/s 4.0 kg 4.0 kg 3.0 kg +x 3.0 kg θ₃ 2.0 m/s top view 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 θ₃ measured clockwise from the +x-direction, as indicated, while the 4.0-kg puck moves at an unknown speed and at an unknown angle θ₄ 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 momentum in x-direction

Before collision, total momentum in x-direction: \( m_1v_{1i} + m_2v_{2i} = 4.0 \times 6.0 + 3.0 \times 0 = 24.0 \, \text{kg·m/s} \)
After collision, total momentum in x-direction: \( m_1v_{1f}\cos\theta_4 + m_2v_{2f}\cos\theta_3 \)
Let \( v_{1f} \) be speed of 4.0 kg puck, \( v_{2f} = 2.0 \, \text{m/s} \)

Step2: Apply conservation of kinetic energy (elastic collision)

Kinetic energy before: \( \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2} \times 4.0 \times 6.0^2 + 0 = 72.0 \, \text{J} \)
Kinetic energy after: \( \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 = \frac{1}{2} \times 4.0 \times v_{1f}^2 + \frac{1}{2} \times 3.0 \times 2.0^2 \)
Set equal: \( 72 = 2v_{1f}^2 + 6 \)
Solve for \( v_{1f} \):
\( 2v_{1f}^2 = 72 - 6 = 66 \)
\( v_{1f}^2 = 33 \)
\( v_{1f} = \sqrt{33} \approx 5.7 \, \text{m/s} \)

Answer:

D. 5.7 m/s