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# Wich will roll the farthest ?

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Hi.

A toy car receives the impact of a horizontally going projectile onto its solid, metallic, vertical rear bumper, and the projectile bounces back.

The same toy, weighing the same as before with a springy or rubber bumper, receives the same projectile and it does not bounce backwards at all, just drops vertically after impact.

Which case will push the car more distance because of the impact ?

Miguel

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I'm not a physicist but i would say the one with the metal bumber presuming all the other factors are the same, as some of the impact of the collision would be absorbed by the rubber bumper.

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The metal, solid car would go further because the ball had more momentum change. Because momentum is conserved, the car also must have a larger momentum change.

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The material of the bumper doesn't make a difference, just the change it invokes.

If the ball bounces back at the same speed it hits the car, but going the other way, it means that the change in momentum is exactly twice that of the change of momentum in the situation where the ball drops strait down.

For example.

momentum = mass * velocity

P = mv

so

ball: 2kg x 2m/s = 4Ns

car: 4kg x 0m/s = 0Ns

When the ball hits the car, and drops strait down, that means that the car receives all 4Ns of momentum, so the car is then moving at 1m/s.

If the ball hits the car and ends up moving at 2m/s in the opposite direction, the change in velocity is not 2m/s, but 4m/s, as it is now moving at -2m/s.

That means the change in momentum is 8Ns, as 2kg x 4m/s = 8Ns.

Bouncy collision:

8Ns/4kg = 2m/s

Non-bouncy collision

4Ns/4kg = 1m/s

In other words, what hurts more:

a bouncy ball to the face, or a mud pie to the face, assuming they weigh the same.

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In other words, what hurts more:

a bouncy ball to the face, or a mud pie to the face, assuming they weigh the same.

No one can possible say as if someone chucked a beach ball at your face or a shot put with the same force one would have far greater implications, also pain would depends on the surface area.

Even if a beach ball had greater velocity it still wouldn't hurt as much, people have hit me straight in the face with one after coming straight off there foot 5 meters before the most you get is a shock because you weren't expecting it.

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Not enough information, BUT...

Based on conservation of energy, the springy bumper car has the potential to go the farthest because all of the kinetic energy was transfered. What is not known is how much energy was transfer from the bumper to the car.

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There is enough information (assuming the cars are of equal mass) because the ball that bounces backward has a much greater change in momentum. Since momentum is conserved, the car will also have a greater change in momentum and since mass is constant this means the velocity increases.

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No one can possible say as if someone chucked a beach ball at your face or a shot put with the same force one would have far greater implications' date=' also pain would depends on the surface area.

Even if a beach ball had greater velocity it still wouldn't hurt as much, people have hit me straight in the face with one after coming straight off there foot 5 meters before the most you get is a shock because you weren't expecting it.[/quote']

We are talking about two objects that are the same size, shape and have the same mass.

If you think a beachball and mud fall into that category together, I'm afraid that's your problem.

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Alt_f13 raises a valid point about conservation of momentum. So, I agree with his answer. The situation where the ball bounces back implies a negative impulse for the ball, so the momentum will be larger for the car in the bounce-back situation.

However, the example alt_f13 supplied is not really correct. There also is conservation of energy. In his example the ball bounces back with the same speed as it had before, so the kinetic energy of the ball would be the same as before. Besides that, the car moves forward and that also has a certain energy. So, the speed of the ball bouncing back will always be smaller than the speed at which it was moving towards the car before impact.

The whole concept of conservation of momentum only is true in ideal collisions with colliding objects having no friction-coupling with an external reference frame. The following example can make things clear:

Suppose the car is fixed to the ground by means of a big pin. It cannot move at all. You may also replace the car by a thick rigid wall. Now, if the ball is moving towards the wall, it bounces back. Still, we have conservation of energy, but do we also still have conservation of momentum???

From a practical point of view we would say no. Momentum is reversed. But now looking at it completely theoretically, we still can say yes. Assume that we have two free frictionless objects with mass M1 and mass M2. M1 is the ball, M2 is the car. Now, if we increases mass M2, then we'll see that total momentum remains the same, but because M2 becomes larger, its speed becomes lower and lower. When M2 approaches infinity (the situation of a big rigid wall, attached to a fixed world), then impulse still is conserved, but the part imposed on M2 does not show up anymore, because its speed goes to 0, as M2 goes to infinity. So, purely theoretically, we still have conservation of momentum.

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However, the example alt_f13 supplied is not really correct. There also is conservation of energy. In his example the ball bounces back with the same speed as it had before, so the kinetic energy of the ball would be the same as before. Besides that, the car moves forward and that also has a certain energy. So, the speed of the ball bouncing back will always be smaller than the speed at which it was moving towards the car before impact.

Actually, the post-impact speed of the object is not specified. There is not enough information to determine whether kinetic energy is conserved in that collision. It is definitely not conserved in the other collision.

The whole concept of conservation of momentum only is true in ideal collisions with colliding objects having no friction-coupling with an external reference frame. The following example can make things clear:

Suppose the car is fixed to the ground by means of a big pin. It cannot move at all. You may also replace the car by a thick rigid wall. Now' date=' if the ball is moving towards the wall, it bounces back. Still, we have conservation of energy, but do we also still have conservation of momentum???

From a practical point of view we would say no. Momentum is reversed. But now looking at it completely theoretically, we still can say yes. Assume that we have two free frictionless objects with mass M1 and mass M2. M1 is the ball, M2 is the car. Now, if we increases mass M2, then we'll see that total momentum remains the same, but because M2 becomes larger, its speed becomes lower and lower. When M2 approaches infinity (the situation of a big rigid wall, attached to a fixed world), then impulse still is conserved, but the part imposed on M2 does not show up anymore, because its speed goes to 0, as M2 goes to infinity. So, purely theoretically, we still have conservation of momentum.[/quote']

Right. It's all about defining the system properly. A wall, fixed to the earth, will have forces exerted on it by the earth. Momentum is conserved when the net external force on the system is zero.

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Hi all.

I do not see why the bounced projectile speed matters for the response, but if a figure is preferred as clarification, just say the projectile bounces back at 30% of the speed it had before impact.

And some other detail if needed later; assume there is no thermal loss in the impact onto the springy or rubber bumper (the rubber -nor the projectile- will not heat up upon collision by deformation)

Miguel

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The metal will still go further because the change in direction of the ball required a greater momentum change which also must be observed in the momentum (velocity in this case since mass is constant) change of the car. Momentum is conserved and a greater momentum change in one object requires the same amount of momentum change in the other (some will be transferred to air resistance but it will still be more than the rubber/springy car.)

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And some other detail if needed later; assume there is no thermal loss in the impact onto the springy or rubber bumper (the rubber -nor the projectile- will not heat up upon collision by deformation)

I don't think you can specify that. The collision will not be elastic, so the kinetic energy has to go somehwere.

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When $b$ is the ball and $c$ is the rigid car -

$m_bv_b+m_cv_c=m_bv_b'+m_cv_c'$

=> $m_bv_b=-.3m_bv_b+m_cv_c'$

=> $v_c' = \frac{1.3m_bv_b}{m_c}$

As opposed to the squishy car -

$m_bv_b+m_cv_c=m_bv_b'+m_cv_c'$

=> $m_bv_b=v'(m_b+m_c)$

=> $v' = \frac{m_bv_b}{m_b+m_c}$

If this is correct, then the the rigid car will have a larger velocity. It will not be exactly the number because of air resistence, heat, etc. but it should still be larger than the rubber/absorbant car.

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Hi Swansont.

I said that just because someone may say the kinetic energy from the projectile dissipated into heat at the spring -or rubber- and will not move the car much because of the energy loss.

As in another extreme example, the bumper with a layer of play-dough to avoid the bounce and drop after; someone can say the dough converted energy to heat in the deformation.

JustStuit: Thanks for your calculations. If the projectile bounces back and still has some speed after the impact, isn't because did not deliver all of its kinetic energy ? and thus, the car will not move as much as in the case of the projectile that did not have energy left after the impact because it delivered all of it to the car ?

Miguel

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JustStuit: Happy birthday!

Woelen: Yeh. I messed up a little. I wanted to get the point across that momentum is a vector quantity, and forgot about the car when I was trying to illustrate that. Should have used a wall :S

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The no rubber bumper would go further as the rubber (depending on it's thickness and construction) would absorb some of the energy

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Hi Swansont.

I said that just because someone may say the kinetic energy from the projectile dissipated into heat at the spring -or rubber- and will not move the car much because of the energy loss.

As in another extreme example' date=' the bumper with a layer of play-dough to avoid the bounce and drop after; someone can say the dough converted energy to heat in the deformation.

[/quote']

My point was that you may have overcontrained the problem by that statement. The action of the impinging particle is enough to determine the result, qualitatively, as JustStuit explained very early on, and alt_f13 confirmed.

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The no rubber bumper would go further as the rubber (depending on it's thickness and construction) would absorb some of the energy

No.

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