Man jumping onto a see-saw to launch bricks vertically

[lioric]

Homework Statement

A man of mass 100kg wants to launch a brick into the air using a see saw.
He wants to launch the bricks of mass 5kg, 10 meters vertically in to the air.
The see saw is placed parallel to the ground using a support.
The bricks are placed on one end of the see saw.
The see saw is made of rigid material with a fulcrum in the center.
The see saw beam is 5m long
If the man jumps onto the far end of the seesaw (5 meters from the bricks) from a height of h meters from where he would land on the see saw.
Assume no air resistance, total energy convertion.

a) Find the value of h
b) What would be the value of h if the fulcrum was placed 1m from the bricks
c) what would be the value of h if the fulcrum was placed 1 m from where the man was to land.

Relevant Equations

W=mg
v=u+at
v²=u²+2as
s=ut+1/2at²
F=ma
GPE=mgh
KE=1/2mv²
Moment = force x perpendicular distance
Work done = force x distance moved

I drew a diagram for the a) part
The person is h meters high
So GPE= 100 x 9.8x h
GPE= 980h j
KE = 980h when the person hits the see saw
KE=1/2mv²
980h=0.5 x 5 x v²

Now it v²=u²+2as
For the brick going up to 10m
v = 0
u=?
a=-9.8ms-²
s=10m

u²=2 x 9.8 x 10
u=14m/s

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

980h = 0.5 x 5 x 14²
980h = 490
h=490/980
h=0.5m

Oh crap
I used the mass of bricks for the person

980h = 0.5 x 100 x 14²
980h = 9800
h=9800/980
h=10m

Is this correct?
I assumed that the v for both sides of the see saw is same cause the fulcrum is in the middle

But don't know how to attempt it when the fulcrum is not in the middle as in part b) and c)

Please help
Thank you in advance

 

[PeroK]

Problem Statement: A man of mass 100kg wants to launch a brick into the air using a see saw.
He wants to launch the bricks of mass 5kg, 10 meters vertically into the air.
The see saw is placed parallel to the ground using a support.
The bricks are placed on one end of the see saw.
The see saw is made of rigid material with a fulcrum in the center.
The see saw beam is 5m long
If the man jumps onto the far end of the seesaw (5 meters from the bricks) from a height of h meters from where he would land on the see saw.
Assume no air resistance, total energy convertion.

a) Find the value of h
b) What would be the value of h if the fulcrum was placed 1m from the bricks
c) what would be the value of h if the fulcrum was placed 1 m from where the man was to land.
Relevant Equations: W=mg
v=u+at
v²=u²+2as
s=ut+1/2at²
F=ma
GPE=mgh
KE=1/2mv²
Moment = force x perpendicular distance
Work done = force x distance moved

View attachment 245060

I drew a diagram for the a) part
The person is h meters high
So GPE= 100 x 9.8x h
GPE= 980h j
KE = 980h when the person hits the see saw
KE=1/2mv²
980h=0.5 x 5 x v²

Now it v²=u²+2as
For the brick going up to 10m
v = 0
u=?
a=-9.8ms-²
s=10m

u²=2 x 9.8 x 10
u=14m/s

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

980h = 0.5 x 5 x 14²
980h = 490
h=490/980
h=0.5m

Is this correct?
I assumed that the v for both sides of the see saw is same cause the fulcrum is in the middle

But don't know how to attempt it when the fulcrum is not in the middle as in part b) and c)

Please help
Thank you in advance

Are you saying that the man jumps from only $0.5m$ and reaches $14m/s$ when he hits the see-saw?

What principles are you applying here to decide what happens when the man lands on the see-saw?

 

[lioric]

Are you saying that the man jumps from only $0.5m$ and reaches $14m/s$ when he hits the see-saw?

What principles are you applying here to decide what happens when the man lands on the see-saw?

Oh crap
I used the mass of bricks for the person

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

980h = 0.5 x 100 x 14²
980h = 9800
h=9800/980
h=10m

 

[PeroK]

Oh crap
I used the mass of bricks for the person

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

980h = 0.5 x 100 x 14²
980h = 9800
h=9800/980
h=10m

You are missing something fundamental. If the brick were tied to the see-saw, then the whole system would rotate as a rigid body. But, the see-saw is not a fully rigid body. The brick will be projected into the air faster than the man lands on the other end. Why?

The physical reason is that a shockwave travels along the see-saw. This is similar to hitting a tennis ball with a racket. The ball leaves the racket much faster than you have moved your arm.

In any case, you need to find the physical laws that govern the transfer of energy from the man to the brick in this case. Any ideas? Hint: what two things are conserved?

 

[haruspex]

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

Why?

 

[lioric]

Why?

Using equations of motion it takes initial 14m/s to lift an object up to 10m

 

[lioric]

You are missing something fundamental. If the brick were tied to the see-saw, then the whole system would rotate as a rigid body. But, the see-saw is not a fully rigid body. The brick will be projected into the air faster than the man lands on the other end. Why?

The physical reason is that a shockwave travels along the see-saw. This is similar to hitting a tennis ball with a racket. The ball leaves the racket much faster than you have moved your arm.

In any case, you need to find the physical laws that govern the transfer of energy from the man to the brick in this case. Any ideas? Hint: what two things are conserved?

Conservation of energy and conservation of momentum

 

[haruspex]

Using equations of motion it takes initial 14m/s to lift an object up to 10m

But that's the launch velocity needed for the brick. Why is that the speed of the man on hitting the seesaw? (Not saying it isn't, just querying your assumption.)

 

[PeroK]

Conservation of energy and conservation of momentum

What momentum, specifically?

Be careful: are there any external forces on the system? Apart from gravity on the man and the brick? Is momentum in the up/down direction conserved during the impact of the man and the launching of the brick?

 

[haruspex]

The question is is quite hard to answer if you take the physics of it seriously. Instead, it seems you are expected to rely on this utterly unrealistic assumption: total energy conversion. I interpret that as meaning all the man's lost PE is converted into man's KE, thence into brick KE, and so into brick PE. That's all you need - forget the seesaw.

 

[PeroK]

The question is is quite hard to answer if you take the physics of it seriously. Instead, it seems you are expected to rely on this utterly unrealistic assumption: total energy conversion. I interpret that as meaning all the man's lost PE is converted into man's KE, thence into brick KE, and so into brick PE. That's all you need - forget the seesaw.

I see it differently. The man collides with the see-saw loses some KE, which is transferred to the brick.

I assume we ignore any rotational KE the see-saw gains durng the short-duration collision that launches the brick.

 

[haruspex]

loses some KE

But how much? Seems to me they are saying "all".
Edit: @PeroK has just pointed out to me that what I am suggesting would violate a physical law. So the next option is to interpret full conversion of energy as no loss of mechanical energy. The man gets to keep some.
Combined with that other law that provides enough info for a solution.

 

[lioric]

What momentum, specifically?

Be careful: are there any external forces on the system? Apart from gravity on the man and the brick? Is momentum in the up/down direction conserved during the impact of the man and the launching of the brick?

This is question meant for a lower like grade 10
We can say that the momentum is conserved
Otherwise we'll have to take account a lot of ways energy gets wasted. So you momentum is conserved

Nope no external forces
But to makes things interesting we can say the force exerted by the brick on the could oppose the person's force

 

[lioric]

But that's the launch velocity needed for the brick. Why is that the speed of the man on hitting the seesaw? (Not saying it isn't, just querying your assumption.)

Well I saw a similar question about a contractor launching a brick to his friend

https://www.varsitytutors.com/ap_physics_1-help/gravitational-potential-energy
I used the idea from that
But I guess since the fulcrum is in the middle and we are treating the see saw as a rigid body no energy is wasted into bending the seesaw while over coming the inertia
But mainly the rigid body rotating from the center can have equal speeds equidistant from the pivot

 

[PeroK]

This is question meant for a lower like grade 10
We can say that the momentum is conserved
Otherwise we'll have to take account a lot of ways energy gets wasted. So you momentum is conserved

Nope no external forces
But to makes things interesting we can say the force exerted by the brick on the could oppose the person's force

There is an external force at the fulcrum. You can see this because the man cannot speed up, so his momentum is reduced. And, the brick is launched upwards, which equates to more loss of momentum in the downward direction. The upward force at the fulcrum, therefore, is an external force which gives an impluse to the system (of man, brick and see-saw) in the upwards direction.

This problem may be too advanced for you. You'll have to decide. The key concept you are missing is conservation of angular momentum: in this case about the fulcrum. We choose to look at angular momentum about the fulcrum because the external force at the fulcrum has no effect on this quantity.

 

[lioric]

The question is is quite hard to answer if you take the physics of it seriously. Instead, it seems you are expected to rely on this utterly unrealistic assumption: total energy conversion. I interpret that as meaning all the man's lost PE is converted into man's KE, thence into brick KE, and so into brick PE. That's all you need - forget the seesaw.

Ya as I said this is meant for grade 10s so they haven't gone through the whole complications. They can use momentum, moment, energy and equations of motion.
And we take everything as an isolated system. No air resistance, total energy and momentum conservation.

 

[haruspex]

We can say that the momentum is conserved

As @PeroK has pointed out, there is an impulse from the fulcrum, so linear momentum is not conserved. What is a third conservation law?

we are treating the see saw as a rigid body no energy is wasted into bending the seesaw

Energy loss is not a simple question of rigid or otherwise. Nothing is completely rigid. The seesaw will behave like a spring, though a very stiff one. Energy loss in it depends on its elasticity, not its rigidity.

 

[lioric]

As @PeroK has pointed out, there is an impulse from the fulcrum, so linear momentum is not conserved. What is a third conservation law?

Energy loss is not a simple question of rigid or otherwise. Nothing is completely rigid. The seesaw will behave like a spring, though a very stiff one. Energy loss in it depends on its elasticity, not its rigidity.

Sorry
Bad choice of words
I was referring to the deforming of the see saw
But I can understand

 

[lioric]

What is a third conservation law?

Are you referring to Newton's third law which related to conservation of angular momentum?

 

[haruspex]

Are you referring to Newton's third law which related to conservation of angular momentum?

Yes. What axis should you choose?

 

[lioric]

Yes. What axis should you choose?

Y axis

 

[jbriggs444]

Y axis

In the context of angular momentum, you should understand "axis" in the sense of "axis of rotation", not in the sense of a coordinate axis.

 

[lioric]

Personally, I would attack the problem by considering the see saw with a fulcrum in the middle as an ideal massless energy-conserving force reverser. Except for the reversal of direction, it is exactly as if the falling man had collided with the brick through an ideal massless energy-conserving spring. Both rebound from what amounts to an elastic collision.

Yes that is exactly how I want this question to be attempt
By the children

So what your saying is I need to put momentum into this and treat it like a collision.
How should I try out the part with the fulcrum not in the middle?

 

[lioric]

In the context of angular momentum, you should understand "axis" in the sense of "axis of rotation", not in the sense of a coordinate axis.

So I guess the action of the man jumping on the seesaw would cause a clockwise moment about the fulcrum

 

[haruspex]

So I guess the action of the man jumping on the seesaw would cause a clockwise moment about the fulcrum

Right, but think about angular momentum about that axis. Would it be conserved and why or why not?

 

[lioric]

Right, but think about angular momentum about that axis. Would it be conserved and why or why not?

The man jumping down on to the seesaw would have an angular momentum of
Mass x velocity (angular) x the distance from the fulcrum.
When the brick leaves the other side the angular momentum is conserved

 

[jbriggs444]

The man jumping down on to the seesaw would have an angular momentum of
Mass x velocity (angular) x the distance from the fulcrum.
When the brick leaves the other side the angular momentum is conserved

Keep talking. Angular momentum is conserved.

What does this mean about the brick's change in angular momentum compared to the man's change in angular momentum?

 

[lioric]

Keep talking. Angular momentum is conserved.

What does this mean about the brick's change in angular momentum compared to the man's change in angular momentum?

Since the man has a higher Mass and the brick is 1/20 the mass of the man, and the distance is the same for both bodies from the fulcrum, the angular momentum will be conserved by the brick having a x20 more velocity

 

[jbriggs444]

Since the man has a higher Mass and the brick is 1/20 the mass of the man, and the distance is the same for both bodies from the fulcrum, the angular momentum will be conserved by the brick having a x20 more velocity

Be careful. It is not twenty times the velocity. It is twenty times the change in velocity.

Can you write that down as an equation?

To figure out how much the two velocities change we are going to need another equation. There are two obvious choices. One assumes the equivalent of an inelastic collision. The other assumes the equivalent of an elastic collision.

Can you pick one of those assumptions and write down a corresponding equation?

 

[lioric]

Be careful. It is not twenty times the velocity. It is twenty times the change in velocity.

Can you write that down as an equation?

To figure out how much the two velocities change we are going to need another equation. There are two obvious choices. One assumes the equivalent of an inelastic collision. The other assumes the equivalent of an elastic collision.

Can you pick one of those assumptions and write down a corresponding equation?

Perfect elastic since KE is conserved

 

[jbriggs444]

Perfect elastic since KE is conserved

OK. Now can you write down some equations? One for conservation of angular momentum and one for conservation of energy?

 

[lioric]

OK. Now can you write down some equations? One for conservation of angular momentum and one for conservation of energy?

Brick from height of h= 10m
PE = mgh = √(KE /0.5 x m)= v of brick

Man with unknown v
Vbrick x distance x mass of brick / Distance x mass of man = v of man

We can use
v²=u² + 2as to find the height of the man jumping off

 

[kuruman]

OK. Now can you write down some equations? One for conservation of angular momentum and one for conservation of energy?

Can I ask why angular momentum is conserved? The rod is made of rigid material which I interpret to mean that if the man end is moving down with instantaneous speed $v$, the brick end is moving up with the same speed. The magnitude of the linear acceleration is the same at both ends for the same reason. Now when the man end hits the ground and stops, the upward speed at the brick end drops to zero very very rapidly meaning an acceleration pointing down starting from an initial value of zero. At the point where the magnitude of the brick end's increasing downward acceleration becomes greater than g, the bricks lose contact and are launched in the air. It is, therefore, safe to assume that the launching speed of the bricks is the same as the speed of the man end just before it hits the ground. In the preceding time interval, after the man lands on the end of the seesaw and before that end hits the ground, angular momentum is not conserved because of the external torque of gravity about the fulcrum.

 

[lioric]

Can I ask why angular momentum is conserved? The rod is made of rigid material which I interpret to mean that if the man end is moving down with instantaneous speed $v$, the brick end is moving up with the same speed. The magnitude of the linear acceleration is the same at both ends for the same reason. Now when the man end hits the ground and stops, the upward speed at the brick end drops to zero very very rapidly meaning an acceleration pointing down starting from an initial value of zero. At the point where the magnitude of the brick end's increasing downward acceleration becomes greater than g, the bricks lose contact and are launched in the air. It is, therefore, safe to assume that the launching speed of the bricks is the same as the speed of the man end just before it hits the ground. In the preceding time interval, after the man lands on the end of the seesaw and before that end hits the ground, angular momentum is not conserved because of the external torque of gravity about the fulcrum.

This is what I wrote at the beginning of this post
Now even I'm confused

 

[jbriggs444]

Can I ask why angular momentum is conserved?

Because the hinge exerts no torque.

Because we are not given the height of the height of the see saw and make the simplifying assumption that its height is negligible. Hence the duration of the interval between impact and touchdown is negligible.

Alternately, because we have assumed an elastic board, there is a bounce and the brick departs the board before the board hits the ground. The torque due to gravity over the interval between impact and the man's touchdown on the ground is irrelevant.

Because the torque from the impact of the man on the ground is irrelevant to the brick. We can end the scenario before it intrudes.

The rod is made of rigid material which I interpret to mean that if the man end is moving down with instantaneous speed vv, the brick end is moving up with the same speed.

That is an inappropriate simplification. You are trying to assume an inelastic interaction in which the speeds match. The OP is positing an elastic interaction.

One can have a rod that is arbitrarily stiff but which remains completely elastic. The limiting behavior of a completely rigid rod is indeterminate unless an additional elasticity parameter is specified.

 

[lioric]

Thank you all very much
I believe I have learned a great deal from this.
This question is something I m making for my students to work on.
May I finish the question properly and post it for you guys to try it out before I give it to my students?

 

[jbriggs444]

May I finish the question properly and post it for you guys to try it out before I give it to my students?

Certainly.

 

[jbriggs444]

Brick from height of h= 10m
PE = mgh = √(KE /0.5 x m)= v of brick

This is extremely sloppy. As written, the above equation states that a velocity is equal to an energy.

What I expect that you mean to have written would equate the final PE of the brick with the initial KE of the brick. e.g. $mgh=\frac{1}{2}mv^2$. Solving for the initial velocity of the brick, one would then get $v=\sqrt{2gh}$.

I also have a pet peeve about writing down variable names without having defined them first. But I'll forgive that in the name of brevity.

Man with unknown v
Vbrick x distance x mass of brick / Distance x mass of man = v of man

Here you seem to be equating the angular momentum of the brick with the angular momentum of the man. That is exactly what I had warned you NOT to do.

The physical meaning of such an equation would be that the interaction between man, brick and seesaw is such that the man stops in place immediately after impact [and before hitting the ground!] and that his prior angular momentum is transferred entirely to the brick.

Such an interaction results in a gain of kinetic energy as a result of the collision. The only way this works is if there is an explosive device under the man's shoes or if he pushes off very strongly in order to bring himself to an immediate stop.

Since you had suggested that mechanical energy is conserved, what I had in mind was equating the initial kinetic energy of the man with the final kinetic energy of man and brick combined.

Conservation of mechanical energy:
$$\frac{1}{2}MV_i^2=\frac{1}{2}MV_f^2 + \frac{1}{2}mv_f^2$$

If you want to go with an inelastic collision then you can apply a simpler equation:

Lever law:
$$v_f=V_f\frac{d_{brick}}{d_{man}}$$
Where V_f is the velocity of the man post-impact.

Either way you can then factor in conservation of angular momentum:
$$mv_fd_{brick} + MV_fd_{man} = MV_id_{man}$$

Where $v_f$ and $V_f$ are the velocities of brick and man post collision and $V_i$ is the velocity of the man post-fall but pre-collision.

We can use
v²=u² + 2as to find the height of the man jumping off

Yes, indeed. Though you might want to re-think the use of the same variable ("v") for both brick and man.

 

[lioric]

Certainly.

Thank you very much
It was truly a delight to work with such brilliant minds

 

[PeroK]

Can I ask why angular momentum is conserved? The rod is made of rigid material which I interpret to mean that if the man end is moving down with instantaneous speed $v$, the brick end is moving up with the same speed. The magnitude of the linear acceleration is the same at both ends for the same reason. Now when the man end hits the ground and stops, the upward speed at the brick end drops to zero very very rapidly meaning an acceleration pointing down starting from an initial value of zero. At the point where the magnitude of the brick end's increasing downward acceleration becomes greater than g, the bricks lose contact and are launched in the air. It is, therefore, safe to assume that the launching speed of the bricks is the same as the speed of the man end just before it hits the ground. In the preceding time interval, after the man lands on the end of the seesaw and before that end hits the ground, angular momentum is not conserved because of the external torque of gravity about the fulcrum.

 

[kuruman]

This is what I wrote at the beginning of this post
Now even I'm confused

Sorry if contributed to your confusion; I had a different model in mind.

 

[lioric]

Sorry if contributed to your confusion; I had a different model in mind.

It's ok. Because of your question I now know when the angular momentum is conserved and when it is not.
I love learning stuff here.
See you guys soon with the final draft of this question