Impulse/ Velocity 2

Question

wo objects travel towards each other and undergo a perfectly elastic interaction. Before they collide, object A has a mass of 373g and a velocity of 9.4m/s, while object B has a mass of 601g and a velocity of 1.1m/s. After they collide, object A travels -0.84292m/s. The interaction takes 0.44s from start to finish.

What is the final velocity of B?    unit   

What is the impulse by A on B?    unit   

What is the force by A on B?    unit   

What is the impulse by B on A?    unit   

What is the force by B on A?    unit   

Answer 1

We can solve this problem using the conservation of momentum and the conservation of kinetic energy.

First, we can calculate the initial momentum and kinetic energy of the system:

Initial momentum = mass A x velocity A + mass B x velocity B = 0.373 kg x 9.4 m/s + 0.601 kg x (-1.1 m/s) [Note that we use a negative velocity for object B since it is moving in the opposite direction to object A] = 2.743 kg m/s

Initial kinetic energy = (1/2) x mass A x (velocity A)^2 + (1/2) x mass B x (velocity B)^2 = (1/2) x 0.373 kg x (9.4 m/s)^2 + (1/2) x 0.601 kg x (1.1 m/s)^2 = 18.43 J

Since the collision is perfectly elastic, the total momentum and kinetic energy of the system are conserved:

Total momentum after collision = mass A x velocity A' + mass B x velocity B' where velocity A' and velocity B' are the final velocities of objects A and B, respectively.

Total kinetic energy after collision = (1/2) x mass A x (velocity A')^2 + (1/2) x mass B x (velocity B')^2

We can use these two equations to solve for the final velocity of object B and the impulses and forces involved in the collision.

1. Final velocity of B: Conservation of momentum gives us: 2.743 kg m/s = 0.373 kg x (-0.84292 m/s) + 0.601 kg x velocity B' Solving for velocity B', we get: velocity B' = 3.697 m/s

Therefore, the final velocity of object B is 3.697 m/s.

1. Impulse by A on B: The impulse is defined as the change in momentum, which is given by: Impulse = mass B x (velocity B' - velocity B) = 0.601 kg x (3.697 m/s - (-1.1 m/s)) = 3.54 Ns

Therefore, the impulse by object A on object B is 3.54 Ns.

1. Force by A on B: The force is defined as the rate of change of momentum, which is given by: Force = Impulse / time = 3.54 Ns / 0.44 s = 8.05 N

Therefore, the force by object A on object B is 8.05 N.

1. Impulse by B on A: Since the collision is perfectly elastic, the impulse by object B on object A is equal in magnitude to the impulse by object A on object B, but opposite in direction. Therefore, the impulse by object B on object A is also 3.54 Ns.
2. Force by B on A: Similarly, the force by object B on object A is equal in magnitude to the force by object A on object B, but opposite in direction. Therefore, the force by object B on object A is also 8.05 N.