Two circular pucks on a frictionless, level surface collide off-center. Each has a radius of r = 5.0[cm] and the only forces exerted in the collision are normal to the edges. Puck 1 has a mass of 0.15[kg] and puck 2 has a mass of 0.20[kg]. At the moment of contact, puck 1 has an initial velocity of 0.1732[m/s] to the right and puck 2is at rest. The impact parameter, b = 1.5r.
The Correct Answer and Explanation is:
In this scenario, we are dealing with a collision between two circular pucks on a frictionless surface, and the problem provides key information about their motion, masses, and radii. The goal is to determine the outcome of this collision, which involves both linear and angular momentum.
Given Data:
- Radius of pucks, rrr = 5.0 cm = 0.05 m
- Mass of puck 1, m1m_1m1 = 0.15 kg
- Mass of puck 2, m2m_2m2 = 0.20 kg
- Initial velocity of puck 1, v1v_1v1 = 0.1732 m/s (to the right)
- Initial velocity of puck 2, v2v_2v2 = 0 m/s (at rest)
- Impact parameter, b=1.5r=1.5×0.05=0.075mb = 1.5r = 1.5 \times 0.05 = 0.075 mb=1.5r=1.5×0.05=0.075m
Key Concepts:
- Impact Parameter:
The impact parameter bbb defines the perpendicular distance between the center of mass of puck 1 and the line of action of the velocity vector of puck 1 at the point of contact. Since b=1.5rb = 1.5rb=1.5r, it means the collision is off-center, implying that the pucks will exert forces on each other that lead to both translational and rotational motion. - Conservation of Momentum:
Since the surface is frictionless and no external forces are involved, the total linear momentum of the system will be conserved. The initial linear momentum is given by the momentum of puck 1, as puck 2 is initially at rest. The conservation of linear momentum in the horizontal direction: m1v1=m1v1′+m2v2′m_1 v_1 = m_1 v_1′ + m_2 v_2’m1v1=m1v1′+m2v2′ where v1′v_1’v1′ and v2′v_2’v2′ are the final velocities of pucks 1 and 2, respectively. - Conservation of Angular Momentum:
Since the force during the collision is normal to the edges of the pucks, each puck will also experience rotational motion. The conservation of angular momentum about the center of mass of the system must be considered. The initial angular momentum is due to the off-center nature of the collision. The initial angular momentum of puck 1 about puck 2’s center of mass is given by: Linitial=m1v1bL_{\text{initial}} = m_1 v_1 bLinitial=m1v1b where bbb is the impact parameter. - Energy Considerations:
In an elastic collision, both kinetic energy and momentum are conserved. However, in an inelastic collision (like the one implied here), only momentum is conserved. We would also need to calculate the velocities of both pucks after the collision, but the problem doesn’t provide enough details on the nature of the collision (perfectly elastic or inelastic). For simplicity, if we assume a perfectly elastic collision, we could calculate the final velocities of the pucks by solving the equations for both momentum and energy conservation.
Conclusion:
To fully solve this problem, we would need to consider both the translational and rotational motion of the pucks and apply conservation laws. By solving the linear and angular momentum equations simultaneously, we can determine the final velocities and rotational speeds of the pucks after the collision.
