TBR Physics #17 momentum

[MrNeuro]

Passage 3 Question #17 page 204

Experiment 2

A block of wood is attached to a long, inelastic string attached to a ceiling. A rubber bullet is fired at the center of the block of wood. In both cases, the block goes into harmonic oscillation after the collision. The two bullets have same mass, size, and speed.

the question is

For experiment 2, how much the height of the block changes depends on:
I. the initial velocity of the bullet
II. the mass of the block.
III. the length of the string

i know that I and II are true. I is due to conservation of energy. II conservation of momentum

however TBR states that the length of the string will not affect the change in height of the block but will affect the height.

Heres where i disagree the bullet is exerting some type of average force F which can be derived from Impulse.

If we were to increase the length of the string then the greater the path/circumference of the harmonic oscillation as circus = 2piR. From this I'm pretty sure this means that the block will have to travel a longer distance in order to reach the same change in height.

so i guess my question is how does the length of the string NOT affect the change in height?

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[Pisiform]

I think this is one of the classical momentum questions in which speed with which block moves depends on x = sqrt 2gh. So this shows that height attained depends only on acceleration due to gravity and velocity of the block. So change in height would be the same imo, no matter what the length is, but as you said, since r increases circumference increases, so arc distance would increase - but height attained would still be the same.

 

[MrNeuro]

I think this is one of the classical momentum questions in which speed with which block moves depends on x = sqrt 2gh. So this shows that height attained depends only on acceleration due to gravity and velocity of the block. So change in height would be the same imo, no matter what the length is, but as you said, since r increases circumference increases, so arc distance would increase - but height attained would still be the same.

x = sqrt 2gh??? or do you mean v = sqrt 2gh ? can't you only assume that if its an elastic collision?

 

[MedPR]

x = sqrt 2gh??? or do you mean v = sqrt 2gh ? can't you only assume that if its an elastic collision?

v=sqrt 2gh is for free-fall.

 

[Pisiform]

x = sqrt 2gh??? or do you mean v = sqrt 2gh ? can't you only assume that if its an elastic collision?

It is an inelastic collision as far as I think. What I did is that, after the bullet hits the block, the block will move with some velocity which i said 'x' (instead of v, because i saved v for velocity of the bullet) and so since the bullet will be embedded in the block, the mass would be mass of the block + mass of the bullet. So KE of the block would be 0.5 (M+m)x^2. This KE would become PE as the block moves and will cause the change in height. So

0.5(M+m)x^2 = (M+m)gH

x = sqrt 2gh

or u can say v = sqrt 2gh ... where v is the speed of the block.

v=sqrt 2gh is for free-fall.

and no not necessarily. Even in fluids you would come across the same equation when youu do question of water falling out of a reservoir. So it depends. There is no hard and fast rule.

 

[MedPR]

It is an inelastic collision as far as I think. What I did is that, after the bullet hits the block, the block will move with some velocity which i said 'x' (instead of v, because i saved v for velocity of the bullet) and so since the bullet will be embedded in the block, the mass would be mass of the block + mass of the bullet. So KE of the block would be 0.5 (M+m)x^2. This KE would become PE as the block moves and will cause the change in height. So

0.5(M+m)x^2 = (M+m)gH

x = sqrt 2gh

or u can say v = sqrt 2gh ... where v is the speed of the block.



and no not necessarily. Even in fluids you would come across the same equation when youu do question of water falling out of a reservoir. So it depends. There is no hard and fast rule.

Oh, I thought the v=sqrt 2gh equation was derived from the kinematics equations when finding the final velocity from free fall, or any situation in which voy is 0 and the only force acting is gravity. Which, in your example, holds true for water falling out of a reservoir.

 

[dranet]

Without seeing the whole story, experimentally it is always found that changing the length of the string only affects the Period and the Frequency of oscillations. Firing a bullet into the block will change the total energy of the system by adding to it in two ways. It adds mass to the block and velocity. Its one of those questions that seem obvious because you think you can imagine that happening. But when you do the expt the length of the string does not affect the height the mass will ascend to.

 

[MrNeuro]

It is an inelastic collision as far as I think. What I did is that, after the bullet hits the block, the block will move with some velocity which i said 'x' (instead of v, because i saved v for velocity of the bullet) and so since the bullet will be embedded in the block, the mass would be mass of the block + mass of the bullet. So KE of the block would be 0.5 (M+m)x^2. This KE would become PE as the block moves and will cause the change in height. So

0.5(M+m)x^2 = (M+m)gH

x = sqrt 2gh

or u can say v = sqrt 2gh ... where v is the speed of the block.



and no not necessarily. Even in fluids you would come across the same equation when youu do question of water falling out of a reservoir. So it depends. There is no hard and fast rule.

if it is indeed an elastic collision as you say it is then you can't use conservation of mechanical energy as it is NOT conserved....

so it would have to be

.5(M)v^2=Mgh

but still how does v= sqrt 2gh prove that the block has the same change in height?

 

[MT Headed]

but still how does v= sqrt 2gh prove that the block has the same change in height?

Because h=(v^2)/(2g)

You could change the height by changing the value of Vfinal (which due to the conservation of momentum only depends on the mass of the bullet, the mass of the block, and the Vinitial of the bullet). Or you could change the height by changing the value of 'g' by running the experiment on a mountain top (or on the moon). It doesn't depend on anything else.

 

[MedPR]

if it is indeed an elastic collision as you say it is then you can't use conservation of mechanical energy as it is NOT conserved....

so it would have to be

.5(M)v^2=Mgh

but still how does v= sqrt 2gh prove that the block has the same change in height?

I thought the definition of an elastic collision is one in which momentum and energy are conserved. Inelastic collisions are conservation of momentum only. Am I backwards?

 

[MT Headed]

No, neuro is backwards.

 

[FishyTheFish]

There are two ways to approach this problem, one being a more logical approach, the other a more mathematical approach.

The logical approach is simple. The only force field acting in this system is gravity, which only works in the y-direction. When the bullet hits the block, and the block starts to swing in a arc like fashion, the distance it travels in the x-direction can be neglected because no energy is required to make it move in the x-direction. However, work is required to move it in the y-direction and changing the length of the string only changes the distance moved in the x-direction, because the y-direction distance moved must remain the same from conservation of energy. This extra distance traversed in the x-direction is not a violation of conservation of energy because there is no force-field acting in the x-direction and thus no energy is required to make the block or bullet move further in the x-direction.


The mathematical/physics approach is as follows:

mv = (M + m)x by conservation of momentum: where m is the mass of the bullet, v is the velocity, M is the mass of the block and x is the velocity of the bullet + block after the collision.

x = mv/(M + m) Solve for the velocity of the bullet + block combo after the collision.

1/2*m*x^2 = (M + m)*g*h After the collision, conservation of energy can be used (The kinetic energy of the bullet + block combo at the bottom of the swing will equal the potential energy at the top of the swing).

Now just solve for h, the height of the bullet + block combo.

1/2*m^3*v^2/(M + m)^2 = (M + m) *g*h

1/2*m^3*v^2 = (M + m)^3*g*h

and so h = (m^3*v^2)/[2*(M + m)^3*g] clearly h only depends on the mass of the bullet, block, and velocity of the bullet. The length of the string is not a factor.

If you need me to clarify, just let me know.

 

[MrNeuro]

I thought the definition of an elastic collision is one in which momentum and energy are conserved. Inelastic collisions are conservation of momentum only. Am I backwards?

No, neuro is backwards.

my bad i meant inelastic...pisiform wrote that its inelastic and then used conservation of energy...

 

[MrNeuro]

There are two ways to approach this problem, one being a more logical approach, the other a more mathematical approach.

The logical approach is simple. The only force field acting in this system is gravity, which only works in the y-direction. When the bullet hits the block, and the block starts to swing in a arc like fashion, the distance it travels in the x-direction can be neglected because no energy is required to make it move in the x-direction. However, work is required to move it in the y-direction and changing the length of the string only changes the distance moved in the x-direction, because the y-direction distance moved must remain the same from conservation of energy. This extra distance traversed in the x-direction is not a violation of conservation of energy because there is no force-field acting in the x-direction and thus no energy is required to make the block or bullet move further in the x-direction.


The mathematical/physics approach is as follows:

mv = (M + m)x by conservation of momentum: where m is the mass of the bullet, v is the velocity, M is the mass of the block and x is the velocity of the bullet + block after the collision.

x = mv/(M + m) Solve for the velocity of the bullet + block combo after the collision.

1/2*m*x^2 = (M + m)*g*h After the collision, conservation of energy can be used (The kinetic energy of the bullet + block combo at the bottom of the swing will equal the potential energy at the top of the swing).

Now just solve for h, the height of the bullet + block combo.

1/2*m^3*v^2/(M + m)^2 = (M + m) *g*h

1/2*m^3*v^2 = (M + m)^3*g*h

and so h = (m^3*v^2)/[2*(M + m)^3*g] clearly h only depends on the mass of the bullet, block, and velocity of the bullet. The length of the string is not a factor.

If you need me to clarify, just let me know.

hmm by the looks of your mathematical approach it looks like your assuming its an inelastic collision and by that then you cannot use conservation of energy right?

how do you justify using conservation of energy in an inelastic collision?

 

[MrNeuro]

Because h=(v^2)/(2g)

You could change the height by changing the value of Vfinal (which due to the conservation of momentum only depends on the mass of the bullet, the mass of the block, and the Vinitial of the bullet). Or you could change the height by changing the value of 'g' by running the experiment on a mountain top (or on the moon). It doesn't depend on anything else.

ahhh this makes a lot more sense....now what if the collision is inelastic

you can no longer use conservation of energy, and can only use conservation of momentum so then how would you determine the change in height using conservation of momentum to find the final velocity and then apply that to v^2 = u^2 + 2as (v^2/2g=h)?

 

[Pisiform]

First of all,

Elastic Collision: Momentum is conserved, KE is conserved, Total Energy is Conserved

Inelastic Collision: Momentum is conserved, KE is NOT conserved, Total Energy is STILL conserved.

Since it is the inealstic collision, we will find out the velocity with which the block will after collision by setting up an equation of:- momentum before collision = momentum after collision.

Once we know the velocity with which the block will move, we can find its KE by just doing 0.5m_blockv^2. Now during the swinging motion, the box will come at rest just like in pendulum ... so its KE would be converted to PE. Equate that and you will get your answer,

also note that I did not do: 0.5m_bullet v^2_bullet = 0.5 m_blockv^2_block ===> because KE is not conserved. I found out the KE using conservation of momentum. And once the block is moving, it is done with collision and your problem now becomes a simple energy one.

 

[MrNeuro]

First of all,

Elastic Collision: Momentum is conserved, KE is conserved, Total Energy is Conserved

Inelastic Collision: Momentum is conserved, KE is NOT conserved, Total Energy is STILL conserved.

Since it is the inealstic collision, we will find out the velocity with which the block will after collision by setting up an equation of:- momentum before collision = momentum after collision.

Once we know the velocity with which the block will move, we can find its KE by just doing 0.5m_blockv^2. Now during the swinging motion, the box will come at rest just like in pendulum ... so its KE would be converted to PE. Equate that and you will get your answer,

also note that I did not do: 0.5m_bullet v^2_bullet = 0.5 m_blockv^2_block ===> because KE is not conserved. I found out the KE using conservation of momentum. And once the block is moving, it is done with collision and your problem now becomes a simple energy one.

got it thanks a lot everyone for the clarification.

 

[MrNeuro]

I think this is one of the classical momentum questions in which speed with which block moves depends on x = sqrt 2gh. So this shows that height attained depends only on acceleration due to gravity and velocity of the block. So change in height would be the same imo, no matter what the length is, but as you said, since r increases circumference increases, so arc distance would increase - but height attained would still be the same.

There are two ways to approach this problem, one being a more logical approach, the other a more mathematical approach.

The logical approach is simple. The only force field acting in this system is gravity, which only works in the y-direction. When the bullet hits the block, and the block starts to swing in a arc like fashion, the distance it travels in the x-direction can be neglected because no energy is required to make it move in the x-direction. However, work is required to move it in the y-direction and changing the length of the string only changes the distance moved in the x-direction, because the y-direction distance moved must remain the same from conservation of energy. This extra distance traversed in the x-direction is not a violation of conservation of energy because there is no force-field acting in the x-direction and thus no energy is required to make the block or bullet move further in the x-direction.


The mathematical/physics approach is as follows:

mv = (M + m)x by conservation of momentum: where m is the mass of the bullet, v is the velocity, M is the mass of the block and x is the velocity of the bullet + block after the collision.

x = mv/(M + m) Solve for the velocity of the bullet + block combo after the collision.

1/2*m*x^2 = (M + m)*g*h After the collision, conservation of energy can be used (The kinetic energy of the bullet + block combo at the bottom of the swing will equal the potential energy at the top of the swing).

Now just solve for h, the height of the bullet + block combo.

1/2*m^3*v^2/(M + m)^2 = (M + m) *g*h

1/2*m^3*v^2 = (M + m)^3*g*h

and so h = (m^3*v^2)/[2*(M + m)^3*g] clearly h only depends on the mass of the bullet, block, and velocity of the bullet. The length of the string is not a factor.

If you need me to clarify, just let me know.

so after doing the chapter on pendulums i now know more about this question...heres where i still have an issue

mgh=mgL(1-cos(theta)) hence
h=L(1-cos(theta))

"mgh" = 1/2 m v^2 then
v=sqrt 2gh

v=sqrt2g(L(1-cos(theta)) or h = v^2 / 2g or

L(1-cos(theta)) = v^2/2g

so yes i agree that mass of the objects and velocity of the objects can affect the change in height but according to h=L(1-cos(theta)) so can length of the string. The only way i can envision length not affecting change in heigh is if somehow L cancels out when you do change in height? ahhhh wait i think it does am i right?

 

[BerkReviewTeach]

so after doing the chapter on pendulums i now know more about this question...heres where i still have an issue

mgh=mgL(1-cos(theta)) hence
h=L(1-cos(theta))

"mgh" = 1/2 m v^2 then
v=sqrt 2gh

v=sqrt2g(L(1-cos(theta)) or h = v^2 / 2g or

L(1-cos(theta)) = v^2/2g

so yes i agree that mass of the objects and velocity of the objects can affect the change in height but according to h=L(1-cos(theta)) so can length of the string. The only way i can envision length not affecting change in heigh is if somehow L cancels out when you do change in height? ahhhh wait i think it does am i right?

The fact here is that if L changes, then theta also changes. For instance, if the string is longer, then it doesn't need to sweep across as large of an angle to have the ball achieve the same height. If the string is shorter, then it needs to sweep across a larger angle to have the ball achieve the same height. You can draw triangles to see this, where any change in the length of the hypotenuse must be balanced out by a change in the angles to keep a given side constant in terms of length.

So in that equation, if L increases which causes theta to decrease, then we get a larger L multipled by a smaller value for (1 - cos(theta)), which offset one another.

 

[MrNeuro]

I think this is one of the classical momentum questions in which speed with which block moves depends on x = sqrt 2gh. So this shows that height attained depends only on acceleration due to gravity and velocity of the block. So change in height would be the same imo, no matter what the length is, but as you said, since r increases circumference increases, so arc distance would increase - but height attained would still be the same.

The fact here is that if L changes, then theta also changes. For instance, if the string is longer, then it doesn't need to sweep across as large of an angle to have the ball achieve the same height. If the string is shorter, then it needs to sweep across a larger angle to have the ball achieve the same height. You can draw triangles to see this, where any change in the length of the hypotenuse must be balanced out by a change in the angles to keep a given side constant in terms of length.

So in that equation, if L increases which causes theta to decrease, then we get a larger L multipled by a smaller value for (1 - cos(theta)), which offset one another.

according to pisiform the arc length increases as the rope gets longer....which makes more sense than a small rope having to sweep across a larger angle.

is that true?

or are you saying the length in the y direction? if not then the gist of your post is
that h=L(1-cos(Theta)

as L increases theta decreases =1 - larger number = Large L (small #)
where L decreases theta increases= 1 - smaller number = small L (larger #)

making them both equal...right? if thats true then the only thing I'm still having trouble visualizing is the whole small rope sweeps across larger angle..

btw great to have you back Berk

 

[milski]

Does this help visualizing it?

 

[MrNeuro]

Does this help visualizing it?

so would it be correct to say that the longer string requires a greater change in the Y direction in a shorter amount of time. due to this the angle is greater?

and thanks for the pic really helped

 

[milski]

so would it be correct to say that the longer string requires a greater change in the Y direction in a shorter amount of time. due to this the angle is greater?

and thanks for the pic really helped

I'm not sure that time matters here. If the vertical displacement is the same, the angle needs to be larger when the radius is smaller. One way to think about that is that if the angle was the same, for the smaller radius everything will be proportionally smaller and the vertical displacement will be smaller too.