Kinetic Energy of Ball Thrown by Woman on Train

In summary: The KE gained by the ball, from the point of view of someone on the ground, is (1/2)mu2+ muv. Thank you for your help!In summary, the woman throwing the ball gains kinetic energy of (1/2)mv^2 relative to herself, and (1/2)mu^2 + muv relative to a person standing on the ground. The work done by the woman throwing the ball is (1/2)mv^2, and the work done by the train is mvu. This work can be deduced using conservation of momentum, as the train loses momentum and would slow down without power applied.
  • #1
mindcircus
11
0
A train moves along the tracks at a constant speed u. A woman on the train throws a ball of mass m straight ahead with a speed v with repsect to herself. What is the kinetic energey gain of the ball as measured by a person on the train? By a person standing by the railroad track? How much work is done by the woman throwing the ball and the train?

I'm not completely sure I have these concepts down.

With respect to the woman, wouldn't kinetic energy just be (1/2)mv^2? I think this because the woman, ball, and reference frame are all moving at the same speed, initially. But for someone on the side of the tracks, I know the same logic can't apply. The woman and ball are moving at the speed of the train, while the reference frame is stationary...I'm not quite sure where to go with this.
Work is just the change in kinetic energy. But I don't know how to deduce the work done by the train. The ball is moving with the horizontal speed of the train. Do I just use that speed in the equation of KE? I'm not really sure how to figure this out.

Thank you for your help!
 
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  • #2
Yes, relative to the woman, the kinetic energy gained by the ball will be (1/2) mv2. To a person standing on the ground, the original speed of the ball was u and the new speed was u+v. The change in kinetic energy observed by the person on the ground is still (1/2)mv2 since the speed of the ball has still changed by v. (And (1/2)mv2 is the work done by the woman in either reference frame.)
 
  • #3
Originally posted by HallsofIvy
Yes, relative to the woman, the kinetic energy gained by the ball will be (1/2) mv2. To a person standing on the ground, the original speed of the ball was u and the new speed was u+v. The change in kinetic energy observed by the person on the ground is still (1/2)mv2 since the speed of the ball has still changed by v. (And (1/2)mv2 is the work done by the woman in either reference frame.)

And would this also be the work done by the train? When the woman throws, the force of the throw is translated down her body through her feet to the train. This becomes a rear-ward acceleration equal to the force of the throw. To maintian a constant speed, the train's engine must provide a forward acceleration of equal force, right?

The way the question is phrased, I don't think it could be asking the work done by the train to accelerate the ball from stationary (relative to the otuside observer) to u+v.
 
  • #4
Originally posted by HallsofIvy
Yes, relative to the woman, the kinetic energy gained by the ball will be (1/2) mv2. To a person standing on the ground, the original speed of the ball was u and the new speed was u+v. The change in kinetic energy observed by the person on the ground is still (1/2)mv2 since the speed of the ball has still changed by v. (And (1/2)mv2 is the work done by the woman in either reference frame.)
That doesn't follow. To a person on the ground, the change in kinetic energy of the ball is
[tex]\Delta E_{b,g}={1\over 2}m(v+u)^2-{1\over 2}mu^2={1\over 2}mv^2+mvu[/tex]
The work done by the woman is the same whether observed from the ground or from the train: [itex](1/2)mv^2[/itex]. The extra work done by the train, as pointed out by LURCH, is [itex]mvu[/itex]. You can do this by conservation of momentum. The train loses momentum of [itex]mv[/itex] and so in absence of power applied, would slow down by speed increment [itex]mv/M[/itex], where M is the mass of the train. Convert that to energy lost by the train and we find the missing [itex]mvu[/itex].
 
  • #5
You know, it occurred to me as soon as I clicked the "Submit Reply" button that (u+v)2 is not u2+v2. You are completely correct.
 

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is calculated by the mass of the object multiplied by its velocity squared, and is measured in joules (J).

2. How does a woman throwing a ball on a train affect its kinetic energy?

When the woman throws the ball on a train, she is adding to the ball's initial velocity and therefore increasing its kinetic energy. The train's movement also contributes to the ball's kinetic energy, as it is already in motion. This results in a higher overall kinetic energy for the ball.

3. Does the mass of the ball affect its kinetic energy when thrown by a woman on a train?

Yes, the mass of the ball does affect its kinetic energy. According to the equation for kinetic energy, a higher mass will result in a higher kinetic energy as long as the velocity remains constant. This means that a heavier ball thrown by a woman on a train will have a higher kinetic energy compared to a lighter ball thrown with the same velocity.

4. How does the velocity of the ball change when thrown by a woman on a train?

The velocity of the ball increases when thrown by a woman on a train. This is because the woman's throwing motion adds to the initial velocity of the ball, and the train's movement also contributes to the ball's velocity. This results in a higher overall velocity for the ball compared to if it was thrown while standing still.

5. What are some real-world applications of understanding kinetic energy of a ball thrown by a woman on a train?

Understanding the kinetic energy of a ball thrown by a woman on a train can be useful in various scenarios, such as in sports. For example, in a game of baseball, the pitcher's throwing motion and the movement of the ball after being hit by a batter can be explained using kinetic energy. It can also be useful in engineering, such as in designing roller coasters where the kinetic energy of the cars is an important factor in ensuring a thrilling and safe ride.

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