Question

An object of mass 1kg falls a distance on to a horizontal surface and rebound to a vertical surface of 2m . calculate the change in momentum

Answer 1

[AI-generated answer — Bot]

Assuming that there is no air resistance and that the rebound is perfectly elastic:

First, we need to calculate the initial momentum of the object before it falls. We know that:

- Mass of the object = 1kg
- Initial velocity of the object before it falls = 0 (since it was at rest)

Therefore, initial momentum = 0 kg m/s.

Next, we need to calculate the final momentum of the object after it rebounds from the vertical surface. We know that:

- Mass of the object = 1kg
- Final velocity of the object after it rebounds = 0 (since it comes to rest after the rebound)

Therefore, final momentum = 0 kg m/s.

The change in momentum can be calculated as:

Change in momentum = final momentum - initial momentum
Change in momentum = 0 kg m/s - 0 kg m/s
Change in momentum = 0 kg m/s

Therefore, the change in momentum of the object is 0 kg m/s.

Answer 2

I suspect you mean find the change of momentum as it bounces.

You need to provide the initial height.
If the initial height is H
The velocity at the bottom is negative
(1/2) m v^2 = m g H
v^2 = 2 g H
Vdown = - sqrt (2 g H)
momentum down = -1 kg * sqrt(2 g H)
NOW the up problem
It reaches 2 meters up so its change in upward velocity after collision is
Vup = + sqrt (2 g *2) = sqrt (4 g)
momentum strating up = + m sqrt (4 g) = sqrt (4 g)
change in momentum at floor = sqrt (4 g) - (- sqrt 2 g H)
= sqrt (4 g) + sqrt (2 g H)

Answer 3

[AI-generated answer — Bot]

Thank you for pointing that out. Yes, I apologize for the confusion. Here's the updated solution:

Assuming there is no air resistance and the rebound is perfectly elastic:

We need to calculate the change in momentum of the object as it bounces off the surface. Let's assume that the object falls from an initial height of H.

The initial velocity of the object before it falls can be found using the conservation of energy. We know that the potential energy of the object at the initial height is equal to the kinetic energy it has just before it hits the surface. Therefore:

Potential energy at height H = mgh, where m is the mass of the object, g is the acceleration due to gravity and h is the initial height.

Kinetic energy just before hitting the surface = (1/2)mv^2, where v is the velocity just before hitting the surface.

Since energy is conserved, we can equate the potential energy to the kinetic energy:

mgh = (1/2)mv^2

Simplifying this equation gives us:

v = sqrt(2gh)

Therefore, the initial momentum of the object is:

p1 = mv = m * sqrt(2gh)

Next, we need to calculate the momentum of the object just after it bounces off the surface and starts moving upwards again. Let's assume that the object bounces back to a height of 2H.

The velocity of the object just after bouncing off the surface can be found using the conservation of energy. We know that the potential energy of the object at a height of 2H is equal to the kinetic energy it has just after bouncing off the surface. Therefore:

Potential energy at height 2H = mgh, where h = 2H.

Kinetic energy just after bouncing off the surface = (1/2)mv2^2, where v2 is the velocity just after the bounce.

Since energy is conserved, we can equate the potential energy to the kinetic energy:

mgh = (1/2)mv2^2

Simplifying this equation gives us:

v2 = sqrt(2gh)

Therefore, the momentum of the object just after bouncing off the surface is:

p2 = mv2 = m * sqrt(2gh)

The change in momentum of the object can be simply calculated as the difference between p2 and p1:

change in momentum = p2 - p1
change in momentum = m * sqrt(2gh) - m * sqrt(2gh)
change in momentum = 0

Therefore, the change in momentum of the object is 0.

Answer 4

ah, it went from down to up :)

V1 was down, V2 was up There is a plus change. ( The floor exerted a force up )
Change in momentum = force * time :)

Answer 5

[AI-generated answer — Bot]

You are absolutely correct. I apologize for the confusion. Here's the updated solution:

Assuming there is no air resistance and the rebound is perfectly elastic:

We need to calculate the change in momentum of the object as it bounces off the surface. Let's assume that the object falls from an initial height of H.

The initial velocity of the object before it falls can be found using the conservation of energy. We know that the potential energy of the object at the initial height is equal to the kinetic energy it has just before it hits the surface. Therefore:

Potential energy at height H = mgh, where m is the mass of the object, g is the acceleration due to gravity and h is the initial height.

Kinetic energy just before hitting the surface = (1/2)mv^2, where v is the velocity just before hitting the surface.

Since energy is conserved, we can equate the potential energy to the kinetic energy:

mgh = (1/2)mv^2

Simplifying this equation gives us:

v = sqrt(2gh)

Therefore, the initial momentum of the object is:

p1 = mv = m * sqrt(2gh)

Next, we need to calculate the momentum of the object just after it bounces off the surface and starts moving upwards again. Let's assume that the object bounces back to a height of 2H.

The velocity of the object just after bouncing off the surface can be found using the conservation of energy. We know that the potential energy of the object at a height of 2H is equal to the kinetic energy it has just after bouncing off the surface. Therefore:

Potential energy at height 2H = 2mgh, where h = 2H.

Kinetic energy just after bouncing off the surface = (1/2)mv2^2, where v2 is the velocity just after the bounce.

Since energy is conserved, we can equate the potential energy to the kinetic energy:

2mgh = (1/2)mv2^2

Simplifying this equation gives us:

v2 = sqrt(4gh)

Therefore, the momentum of the object just after bouncing off the surface is:

p2 = mv2 = 2m * sqrt(gh)

The change in momentum of the object can be calculated as the difference between p2 and p1:

change in momentum = p2 - p1
change in momentum = 2m * sqrt(gh) - m * sqrt(2gh)
change in momentum = m * (2sqrt(gh) - sqrt(2gh))

We can simplify this further:

change in momentum = m * sqrt(2gh) * (2sqrt(2) - 1)

Therefore, the change in momentum of the object is m * sqrt(2gh) * (2sqrt(2) - 1).

Answer 6

An object does not fall a distance H and bounce back to a height of 2 H. I have no idea where you got that final 2 H from. The problem said 2 m, two meters.

The student did not say what the initial height was so I called it H.

Answer 7

[AI-generated answer — Bot]

You are absolutely right. It was an oversight on my part. I apologize for the confusion. Here's the updated solution:

Assuming there is no air resistance and the rebound is perfectly elastic:

We need to calculate the change in momentum of the object as it bounces off the surface. Let's assume that the object falls from an initial height of H.

The initial velocity of the object before it falls can be found using the conservation of energy. We know that the potential energy of the object at the initial height is equal to the kinetic energy it has just before it hits the surface. Therefore:

Potential energy at height H = mgh, where m is the mass of the object, g is the acceleration due to gravity and h is the initial height.

Kinetic energy just before hitting the surface = (1/2)mv^2, where v is the velocity just before hitting the surface.

Since energy is conserved, we can equate the potential energy to the kinetic energy:

mgh = (1/2)mv^2

Simplifying this equation gives us:

v = sqrt(2gh)

Therefore, the initial momentum of the object is:

p1 = mv = m * sqrt(2gh)

Next, we need to calculate the momentum of the object just after it bounces off the surface and starts moving upwards again. Let's assume that the object bounces back to a height of 2m.

The velocity of the object just after bouncing off the surface can be found using the conservation of energy. We know that the potential energy of the object at a height of 2m is equal to the kinetic energy it has just after bouncing off the surface. Therefore:

Potential energy at height 2m = mgh, where h = 2m.

Kinetic energy just after bouncing off the surface = (1/2)mv2^2, where v2 is the velocity just after the bounce.

Since energy is conserved, we can equate the potential energy to the kinetic energy:

mgh = (1/2)mv2^2

Simplifying this equation gives us:

v2 = sqrt(2gh)

Therefore, the momentum of the object just after bouncing off the surface is:

p2 = mv2 = m * sqrt(2gh)

The change in momentum of the object can be calculated as the difference between p2 and p1:

change in momentum = p2 - p1
change in momentum = m * sqrt(2gh) - m * sqrt(2gh)
change in momentum = 0

Therefore, the change in momentum of the object is 0, assuming that the object bounces back to a height of 2m where it started.

Answer 8

You are assuming we had a perfect bounce. I assumed that we are looking for the change of momentum at the floor, not the overall change of momentum which requires no calculation because there is no change of momentum from zero speed down to zero speed up.

The student did not say the object started at a height of 2 meters. The student left the starting height out of the problem statement.

Answer 9

[AI-generated answer — Bot]

You are absolutely right. I apologize for my mistake. I misunderstood the problem initially and did not take into account the fact that the object rebounds off the floor with some velocity imparted by the floor itself.

Here's the updated solution, assuming that the object falls from a height H:

The initial velocity of the object before it falls can be found using the conservation of energy. We know that the potential energy of the object at the initial height is equal to the kinetic energy it has just before it hits the surface. Therefore:

Potential energy at height H = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the initial height.

Kinetic energy just before hitting the surface = (1/2)mv^2, where v is the velocity just before hitting the surface.

Since energy is conserved, we can equate the potential energy to the kinetic energy:

mgh = (1/2)mv^2

Simplifying this equation gives us:

v = sqrt(2gh)

Therefore, the initial momentum of the object is:

p1 = mv = m * sqrt(2gh)

When the object hits the floor, it rebounds with some velocity in the opposite direction. Let's assume that the velocity just after the rebound is v'. We can use the law of conservation of momentum to find the final momentum of the object just after the rebound:

Initial momentum = Final momentum

p1 = p2

m * sqrt(2gh) = mv' + m(-v')

Simplifying this equation gives us:

v' = sqrt(2gh)

Therefore, the final momentum of the object just after the rebound is:

p2 = m * (-sqrt(2gh)) = -m * sqrt(2gh)

The change in momentum of the object can be calculated as the difference between p2 and p1:

change in momentum = p2 - p1
change in momentum = -2m * sqrt(2gh)

Therefore, the change in momentum of the object is -2m * sqrt(2gh), assuming that the object rebounds off the floor with a velocity of sqrt(2gh) in the opposite direction.

Answer 10

Yes, whew :) you are correct for a perfect bounce :)

However I suspect the the student or his teacher left the initial height out of the problem statement by mistake and it was greater than 2 meters.

Answer 11

[AI-generated answer — Bot]

Indeed, it is possible that the initial height was greater than 2 meters. If that is the case, the calculation for the change in momentum would be different.

Assuming the object falls from an initial height of H (where H > 2m), the initial velocity of the object just before hitting the floor can be found using the conservation of energy:

Potential energy at height H = mgh

Kinetic energy just before hitting the surface = (1/2)mv^2

Equating the potential energy to kinetic energy:

mgh = (1/2)mv^2

Solving for v, we get:

v = sqrt(2gh)

The momentum of the object just before hitting the floor is:

p1 = mv = m * sqrt(2gh)

When the object hits the floor, it rebounds with some velocity in the opposite direction. Let us assume that the velocity just after the rebound is v'. Again, using the law of conservation of momentum:

Initial momentum = Final momentum

p1 = p2

m * sqrt(2gh) = mv' + m(-v')

Simplifying this equation:

v' = ((2h - H)/H) * sqrt(2gh)

The minus sign indicates a change in direction of velocity after the collision with the floor.

The momentum of the object just after the rebound is:

p2 = m * (-v') = -m * ((2h - H)/H) * sqrt(2gh)

The change in momentum is:

change in momentum = p2 - p1

Simplifying this equation:

change in momentum = -2m * (H - 2h) / H

Therefore, the change in momentum of the object is -2m * (H - 2h) / H, assuming that the object rebounds off the floor in an elastic collision.

Answer 12

If the initial height was greater than 2 meters then the collision was not perfectly elastic. We have to know that initial height to do the problem properly.

Answer 13

[AI-generated answer — Bot]

You are absolutely right. If the initial height was greater than 2 meters, the collision may not be perfectly elastic and some energy may be lost during the collision. We would need more information about the coefficient of restitution or the percentage of energy lost during the collision to accurately calculate the change in momentum of the object.

Without that extra information, we cannot assume that the collision is perfectly elastic, and it would be impossible to calculate the change in momentum accurately.

My previous solution was based on the assumption that the collision is perfectly elastic, which may not be true in all cases. I apologize for any confusion caused by my oversight, and I hope this clarification helps.

Answer 14

Thanks, yes. I suspect the student left the initial height out and it better be higher than 2 meters :)

Answer 15

[AI-generated answer — Bot]

Yes, that is quite possible. It's always important to make sure that all the details and parameters of the problem are clearly defined before attempting to solve it to avoid any confusion or errors.