# Find the height the block rises to after an elastic collision

• Neon32
In summary, two blocks of masses 5.09 kg and 9.50 kg are released from A and B respectively, with a north pole of a strong magnet protruding from the front end of the first block, repelling the north pole of an identical magnet embedded in the back end of the second block. The maximum height to which the first block rises after the elastic collision can be calculated using the equation Vf=(m1-m2/m1+m2)Vi. This method avoids calculating any velocities.
Neon32

## Homework Statement

Two blocks are free to slide along a frictionless wooden track ABC as shown below. The block of mass m1 = 5.09 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 9.50 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

figure:

## Homework Equations

We'll use law of energy conservation: KEi+P.Ei=K.Ef+P.E
+
Vf=(m1-m2/m1+m2)Vi

## The Attempt at a Solution

K.Ei+P.Ei=K.Ef+P.Ef
m1gh=1/2 m1 V1i2[/B]
So we get V1i=9.90 m/s.

Substituting in V1f=(m1-m2/m1+m2)Vi, w get:
V1f=-3.3m/s

1)I want to know if what I did above is correct.

2) My second question is

I'll use the law of energy conservation again so we can find the maximum height:
The second body will acquire velocity so there is kinetic energy after collision:

1/2m1 v12+1/2 m2 V22=m1gh(new height)
Is this one correct?

Last edited:
Updated.

Neon32 said:
m1gh=1/2 m1 V1i2
Did not get this part.
it is given that ##V_{1i} = 0##.

Buffu said:
Did not get this part.
it is given that ##V_{1i} = 0##.

This is velocity of mass 1 right before it collides with body 2.

Neon32 said:
Substituting in V1f=(m1-m2/m1+m2)Vi, w get:
V1f=-3.3m/s

1)I want to know if what I did above is correct.
Yes that is correct. (if the calculations are correct).

Neon32 said:
1/2m1 v12+1/2 m2 V22=m1gh(new height)
You did not clarify what v12 and v22 means here so I assume :-
v12 - velocity of m1 after collision.
v22 - velocity of m2 after collision.

You will get the same height this way (because 1/2m1 v12+1/2 m2 V22 = initial energy of the body m1). Which is incorrect since some energy is transferred this way.
You should not consider the kinetic energy of second body in your equation, i.e 1/2 m2 V22 should not come.

Please use subscript and superscript in writing variables, it is hard to understand.

Buffu said:
Which is incorrect since some energy is transferred this way.
Sorry for this ambiguous statement. I meant to say that some energy is transferred to m2, so height of m1 will be less than initial.

The question looks a bit flawed to me. If the magnets are strong enough to prevent collision, m2 will start to move when m1 is still some way short of it. From the diagram, it will still be on the arc. This means momentum will not be conserved, since the normal force has a horizontal component and it will be more than merely enough to support m1.
Glossing over that...
Neon32 said:
Substituting in V1f=(m1-m2/m1+m2)Vi, w get:
Please use parentheses correctly. That cannot be what you mean.
Neon32 said:
V1f=-3.3m/s
Please check that. I make it a bit less.

Last edited:
BvU
Neon32 said:

## Homework Equations

Vf=(m1-m2/m1+m2)Vi

If you know that equation, can see you a good way to solve the problem without calculating any velocities?

BvU

## 1. What is an elastic collision?

An elastic collision is a type of collision between two objects where there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision remains the same.

## 2. How is the height the block rises to after an elastic collision calculated?

The height the block rises to after an elastic collision can be calculated using the conservation of energy principle. The initial kinetic energy of the block is converted into potential energy as it rises, so we can equate the two energies to find the height. The equation used is: mgh = 1/2mv2, where m is the mass of the block, g is the acceleration due to gravity, and v is the velocity of the block after the collision.

## 3. What factors can affect the height the block rises to after an elastic collision?

The height the block rises to after an elastic collision is affected by the mass and velocity of the block as well as the angle at which it is launched. The surface on which the collision occurs can also play a role, as well as any external forces acting on the block during the collision.

## 4. Is the height the block rises to after an elastic collision the same for all collisions?

No, the height the block rises to after an elastic collision will vary depending on the initial conditions of the collision. This includes the mass and velocity of the block, as well as the angle at which it is launched and the surface on which the collision occurs.

## 5. Can the height the block rises to after an elastic collision be greater than the initial height?

Yes, the height the block rises to after an elastic collision can be greater than the initial height if the initial kinetic energy of the block is greater than the potential energy at the initial height. This can happen if the block is launched at a high enough velocity or if it collides with a surface that can provide an additional upward force, such as a spring.

Replies
10
Views
3K
Replies
4
Views
4K
Replies
10
Views
3K
Replies
5
Views
1K
Replies
17
Views
4K
Replies
5
Views
2K
Replies
4
Views
2K
Replies
1
Views
4K
Replies
2
Views
1K
Replies
1
Views
2K