Question

A train is moving along a track with constant speed v1 relative to the ground. A person on the train holds a ball of mass m and throws it toward the front of the train with a speed v2 relative to the train.
Part A

Calculate the change in kinetic energy of the ball in the Earth frame of reference.

Express your answer in terms of the variables m, v1, and v2.

Part B

Calculate the change in kinetic energy of the ball in the train frame of reference.

Express your answer in terms of the variables m, v1, and v2.

Part C

How much work was done on the ball in the Earth frame of reference?

Express your answer in terms of the variables m, v1, and v2.

Part D

How much work was done on the ball in the train frame of reference?

Express your answer in terms of the variables m, v1, and v2.

Answer 1

Answer:

$(a)0.5m(2v1v2+v2^{2})\\(b)0.5m(v2^{2}-v1^{2})\\(c)0.5m(2v1v2+v2^{2})\\(d)0.5m(v2^{2}-v1^{2})$

Explanation:

velocity of ball in train reference = v2

velocity of ball in earth reference = v1+v2

(a)

Kinetic energy is given by $0.5mv^{2}$ where m and v are the mass and velocity of object respectively.

Change in kinetic energy is given by subtracting initial kinetic energy from the final kinetic energy. In this case

Initial kinetic energy= $0.5mv1^{2}$

Final kinetic energy= $0.5m (v1+v2)^{2}=0.5m(v1^{2}+2v1v2+v2^{2})$

Change in kinetic energy=$0.5m (v1+v2)^{2}-0.5mv1^{2}=0.5m((v1+v2)^{2})-v1^{2}=0.5m(2v1v2+v2^{2}$

(b)

Change in velocity in train reference will be

Initial kinetic energy= $0.5mv1^{2}$

Final kinetic energy= $0.5mv2^{2}$

Change in kinetic energy=$0.5m(v2^{2}-v1^{2})$

(c)

Work done, W = change in kinetic energy=$0.5m (v1+v2)^{2}-0.5mv1^{2}=0.5m((v1+v2)^{2})-v1^{2}=0.5m(2v1v2+v2^{2}$

(d)

Work done, W = change in kinetic energy=$0.5m(v2^{2}-v1^{2})$