Collision w/ zero Energy Loss not making sense

In summary: In a real world scenario, there are many factors that can affect the outcome of a collision, such as friction and air resistance. In the bowling ball and pool ball scenario, the friction between the two balls and the surface they are rolling on would likely prevent the pool ball from bouncing off at a higher speed than the bowling ball's original speed. However, in a hypothetical scenario with no friction or air resistance, it is indeed possible for the pool ball to bounce off at a higher speed. This is because in an elastic collision, both kinetic energy and momentum are conserved, meaning that the total energy and momentum of the system before and after the collision are the same. Therefore, it is possible for the pool ball to gain more kinetic energy from
  • #1
Ocata
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5

Homework Statement



Block m1 has mass 10kg and 10m/s
Block m2 has mass 5kg and 0m/s

Energy loss is 0%

Homework Equations



Conservation of momentum
Conservation of energy
Quadratic Equation

The Attempt at a Solution



(mv)1i + (mv)2i = (mv)1f + (mv)2f

10(10) + 5(0) = 10(v)1f + 5(v)2f

[itex] v_{2f} = 20 - 2(v_{1f})[/itex] Then substituting v2f into the following...

Then

[itex]E_{i} = E_{f}[/itex]

[itex](\frac{1}{2}mv^{2})_{1i} + (\frac{1}{2}mv^{2})_{2i} = (\frac{1}{2}mv^{2})_{1f} + (\frac{1}{2}mv^{2})_{2f}[/itex]

[itex] 1000 = 10v^{2}_{1f} + 5v^{2}_{2f} [/itex]

[itex] 1000 = 10v^{2}_{1f} + 5(20 - 2v_{1f})^{2} [/itex]

[itex] 0 = 30v^{2}_{1f} - 400v_{1f} + 1000[/itex]

by quadratic formula:

[itex] v_{1f} = 10m/s [/itex] or 3.33m/s

Then [itex] v_{2f} = 20 - 2(3.33m/s) = 13.333333[/itex]

or

[itex] v_{2f} = 20 - 2(10m/s) = 0m/s[/itex]

So, the two possible sets of final velocities are:

Possibility 1) velocity of [itex] m_{1f} = 10m/s [/itex] and [itex] m_{2f} = 0m/s[/itex]

or

Possibility 2) velocity of [itex] m_{1f} = 3.333m/s [/itex] and [itex] m_{2f} = 13.333m/s[/itex]Neither case makes sense to me. In the first case, how can something traveling at 10m/s hit a stationary object and continue traveling at 10m/s while the stationary object remains stationary? Maybe I have it mixed up or something? Maybe the object traveling at 10m/s actually becomes stationary while the stationary object is propelled to 10m/s?

In the second case, how can an object traveling at 10m/s collide with an object and cause the object to travel faster than it? Is that even possible?

Thanks
 
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  • #2
Ocata said:

Homework Statement



Block m1 has mass 10kg and 10m/s
Block m2 has mass 5kg and 0m/s

Energy loss is 0%

Homework Equations



Conservation of momentum
Conservation of energy
Quadratic Equation

The Attempt at a Solution



(mv)1i + (mv)2i = (mv)1f + (mv)2f

10(10) + 5(0) = 10(v)1f + 5(v)2f

[itex] v_{2f} = 20 - 2(v_{1f})[/itex] Then substituting v2f into the following...

Then

[itex]E_{i} = E_{f}[/itex]

[itex](\frac{1}{2}mv^{2})_{1i} + (\frac{1}{2}mv^{2})_{2i} = (\frac{1}{2}mv^{2})_{1f} + (\frac{1}{2}mv^{2})_{2f}[/itex]

[itex] 1000 = 10v^{2}_{1f} + 5v^{2}_{2f} [/itex]

[itex] 1000 = 10v^{2}_{1f} + 5(20 - 2v_{1f})^{2} [/itex]

[itex] 0 = 30v^{2}_{1f} - 400v_{1f} + 1000[/itex]

by quadratic formula:

[itex] v_{1f} = 10m/s [/itex] or 3.33m/s

Then [itex] v_{2f} = 20 - 2(3.33m/s) = 13.333333[/itex]

or

[itex] v_{2f} = 20 - 2(10m/s) = 0m/s[/itex]

So, the two possible sets of final velocities are:

Possibility 1) velocity of [itex] m_{1f} = 10m/s [/itex] and [itex] m_{2f} = 0m/s[/itex]

or

Possibility 2) velocity of [itex] m_{1f} = 3.333m/s [/itex] and [itex] m_{2f} = 13.333m/s[/itex]Neither case makes sense to me. In the first case, how can something traveling at 10m/s hit a stationary object and continue traveling at 10m/s while the stationary object remains stationary? Maybe I have it mixed up or something? Maybe the object traveling at 10m/s actually becomes stationary while the stationary object is propelled to 10m/s?

In the second case, how can an object traveling at 10m/s collide with an object and cause the object to travel faster than it? Is that even possible?

Thanks
There's an interesting toy called Newton's Cradle which illustrates the phenomenon you are having trouble grasping:



The energy from the moving ball on one end is transmitted thru the other balls, which remain stationary. The ball at the opposite end, which is initially at rest, is flung away from the rest of the balls. All of these balls in the cradle have identical masses, so the velocities would remain the same. Any ball which had a smaller mass would be flung away at a higher velocity, in order to conserve momentum.
 
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  • #3
Ocata said:

Homework Statement



Block m1 has mass 10kg and 10m/s
Block m2 has mass 5kg and 0m/s

Energy loss is 0%
...

The Attempt at a Solution


...

So, the two possible sets of final velocities are:

Possibility 1) velocity of [itex] m_{1f} = 10m/s [/itex] and [itex] m_{2f} = 0m/s[/itex]

or

Possibility 2) velocity of [itex] m_{1f} = 3.333m/s [/itex] and [itex] m_{2f} = 13.333m/s[/itex]Neither case makes sense to me. In the first case, how can something traveling at 10m/s hit a stationary object and continue traveling at 10m/s while the stationary object remains stationary? Maybe I have it mixed up or something? Maybe the object traveling at 10m/s actually becomes stationary while the stationary object is propelled to 10m/s?

In the second case, how can an object traveling at 10m/s collide with an object and cause the object to travel faster than it? Is that even possible?

Thanks
In the first case, it's clear that both Kinetic Energy and Momentum are conserved. It's just that the collision does not happen. the final conditions are unchanged from the initial conditions.

In the second case: That is exactly what we expect for an elastic collision.
 
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  • #4
I remember seeing that contraption as a kid on an office desk somewhere and thought it was amazing. Even more amazing at this time, now that I have the power of science to begin understanding what's going on there. Thank you.

I was trying to imagine a bowling ball colliding with a pool ball and then visual if the pool ball would actually bounce off the bowling ball at a faster speed than the bowling ball was originally travelling. You are saying it is absolutely possible. So if I go to the bowling alley and pick a 10lb bowling ball and roll it into a pool ball, will I see the pool ball roll away faster than the bowling ball was originally rolling? If it should happen and does not, I'd imagine it would be due to the downward rotation of the bowling ball opposing the forward rolling of the pool ball at the time of impact (like friction)? And the friction of the floor and air?

And hypothetically, if I go into zero gravity with no air resistance and I propel a bowling ball at some speed directly toward a stationary (relative to the bowling ball) pool ball, you are saying I can bet with confidence that the pool ball will bounce forward at a speed greater than the original speed of the bowling ball?

Thank you
 
  • #5
Ocata said:
I remember seeing that contraption as a kid on an office desk somewhere and thought it was amazing. Even more amazing at this time, now that I have the power of science to begin understanding what's going on there. Thank you.

I was trying to imagine a bowling ball colliding with a pool ball and then visual if the pool ball would actually bounce off the bowling ball at a faster speed than the bowling ball was originally travelling. You are saying it is absolutely possible. So if I go to the bowling alley and pick a 10lb bowling ball and roll it into a pool ball, will I see the pool ball roll away faster than the bowling ball was originally rolling? If it should happen and does not, I'd imagine it would be due to the downward rotation of the bowling ball opposing the forward rolling of the pool ball at the time of impact (like friction)? And the friction of the floor and air?

And hypothetically, if I go into zero gravity with no air resistance and I propel a bowling ball at some speed directly toward a stationary (relative to the bowling ball) pool ball, you are saying I can bet with confidence that the pool ball will bounce forward at a speed greater than the original speed of the bowling ball?

Thank you
What happens to the Pins when the bowling ball strikes them?
 
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  • #6
The pins definitely do seem to crash away at a faster velocity than the ball travels down the lane. So this is in fact the case? If so, it can be stated that a object in space, when colliding with a less massive object, will cause the less massive object to always bounce away at a faster velocity than the heavier ball was originally traveling (if both objects move in a linear path and spinning/rotation is not occurring). Correct?
 
  • #7
Ocata said:
The pins definitely do seem to crash away at a faster velocity than the ball travels down the lane. So this is in fact the case? If so, it can be stated that a object in space, when colliding with a less massive object, will cause the less massive object to always bounce away at a faster velocity than the heavier ball was originally traveling (if both objects move in a linear path and spinning/rotation is not occurring). Correct?
Right .

For a head on, elastic collision, the light object willm move away at close to twice the speed, is the mass difference is great enough.

Look into "center of mass".
 
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  • #8
Will do. Thanks.
 

1. How is it possible for a collision with zero energy loss to occur?

According to the principle of conservation of energy, energy cannot be created or destroyed, only transferred from one form to another. Therefore, a collision with zero energy loss would violate this principle. However, it is possible for a collision to result in a negligible energy loss, which can be explained by factors such as friction, heat dissipation, and sound production. These factors may seem insignificant, but they can account for a small amount of energy loss during a collision.

2. Can a collision with zero energy loss be observed in real life?

In theory, a collision with zero energy loss is not possible. However, in practical situations, it is possible for the energy loss to be extremely small and almost imperceptible. For instance, when two objects collide in outer space where there is no air resistance or other external factors, the energy loss may be negligible. However, in everyday situations on Earth, there will always be some form of energy loss during a collision due to various factors such as air resistance and surface imperfections.

3. How do scientists measure the energy loss during a collision?

The energy loss during a collision can be measured using various methods such as the conservation of momentum and the calculation of kinetic energy before and after the collision. These measurements can provide an estimate of the energy lost during the collision, which can then be compared to the total energy of the system before the collision. In cases where the energy loss is extremely small, it may be difficult to measure accurately.

4. What are some real-life examples of collisions with zero energy loss?

As mentioned before, a collision with zero energy loss is not possible in real-life situations. However, there are instances where the energy loss may be so small that it can be considered negligible. For example, when two billiard balls collide on a perfectly smooth and level table, the energy loss may be extremely small due to the lack of friction and other external factors. However, in reality, there will always be some energy loss due to imperfections in the table and the balls themselves.

5. How does the concept of elasticity relate to collisions with zero energy loss?

Elasticity refers to the ability of an object to return to its original shape after being deformed by an external force. In the context of collisions, it is related to the amount of energy that is lost during the collision. In a perfectly elastic collision, there is no energy loss, and the objects involved will bounce off each other with the same speed and kinetic energy as before the collision. However, in most real-life situations, collisions are not perfectly elastic, and there will always be some energy loss, no matter how small.

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