A 0.143 kgkg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.740 m/sm/s. It has a head-on collision with a 0.298 kgkg glider that is moving to the left with a speed of 2.17 m/sm/s. Suppose the collision is elastic.Part AFind the magnitude of the final velocity of the 0.143 kgkg glider.Express your answer in meters per second.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typev0.143kg𝑣0.143kg =nothingm/sm/sSubmitRequest AnswerPart BFind the direction of the final velocity of the 0.143 kgkg glider.to the rightto the leftSubmitRequest AnswerPart CFind the magnitude of the final velocity of the 0.298 kgkg glider.Express your answer in meters per second.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typenothingm/sm/sSubmitRequest AnswerPart DFind the direction of the final velocity of the 0.298 kgkg glider.to the rightto the left

This problem involves the conservation of momentum and kinetic energy because it's an elastic collision.

Part A: The initial momentum of the system is the sum of the momentum of each glider: p_initial = m1v1_initial + m2v2_initial where m1 = 0.143 kg, v1_initial = 0.740 m/s (to the right), m2 = 0.298 kg, v2_initial = -2.17 m/s (to the left, so it's negative).

The final momentum of the system is also the sum of the momentum of each glider: p_final = m1v1_final + m2v2_final

Because momentum is conserved, p_initial = p_final. We can set the two equations equal to each other and solve for v1_final, the final velocity of the 0.143 kg glider.

Part B: The direction of the final velocity of the 0.143 kg glider depends on the sign of v1_final. If v1_final is positive, the direction is to the right. If v1_final is negative, the direction is to the left.

Part C: We can use the conservation of kinetic energy to find v2_final, the final velocity of the 0.298 kg glider. The initial kinetic energy of the system is the sum of the kinetic energy of each glider: KE_initial = 0.5m1v1_initial^2 + 0.5m2v2_initial^2

The final kinetic energy of the system is also the sum of the kinetic energy of each glider: KE_final = 0.5m1v1_final^2 + 0.5m2v2_final^2

Because kinetic energy is conserved, KE_initial = KE_final. We can set the two equations equal to each other and solve for v2_final.

Part D: The direction of the final velocity of the 0.298 kg glider depends on the sign of v2_final. If v2_final is positive, the direction is to the right. If v2_final is negative, the direction is to the left.