# Solving Momentum Problem: Max Height of m1 After Elastic Collision

• Elmon
In summary, two blocks of masses 4.93 kg and 9.60 kg are released from point A on a frictionless wooden track. The block of mass 4.93 kg has a strong magnet attached to it, repelling an identical magnet embedded in the back end of the block of mass 9.60 kg. The two blocks do not touch and the problem is to calculate the maximum height to which the block of mass 4.93 kg rises after an elastic collision. The problem can be solved by conserving energy and momentum in three steps: the fall of the block, the collision, and the rise of the block. By equating the energies before and after the collision, the maximum height can be calculated.
Elmon
Two blocks are free to slide along a frictionless wooden track ABC as shown in Figure P9.20. The block of mass m1 = 4.93 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 = 9.60 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

The figure shows m1 on a curved ramp at a height of 5 m.

Since it is elastic, I know energy and momentum are conserved. So I have:

(1/2)m1*v1o^2+(1/2)m2*v2o^2 = (1/2)m1*v1f^2+(1/2)m2*v2f^2
and
m1*v1o+m2*v2o = m1*v1f+m2*v2f

m2 is initially at rest, so v2o=0. Now I am not sure how I am supposed to use these to find height, or anything at all for that matter. Can anyone give me a point in the right direction?

Two equations, two unknowns. One of those unknowns will give you the energy of m1 immediately following the collision. Mechanical energy is conserved.

Well, mechanical energy is KE+PE, so

(1/2)m1*v1f+m1*g*hf = (1/2)m1*v1o+m1*g*ho

But I am not sure how to connect that to the other equations I have.

I am having a hard time getting my head around this one.

Since you didn't include the figure, I'm just guessing as to what it shows.

Elmon said:
Well, mechanical energy is KE+PE, so

(1/2)m1*v1f+m1*g*hf = (1/2)m1*v1o+m1*g*ho
Make sure you square those speeds in the KE terms. I assume v1o is the initial speed of m1 immediately after the collision. (That speed is called v1f in your collision equations.)

But I am not sure how to connect that to the other equations I have.
The collision equations will give you the speed of m1 after the collision.

There are three steps to this problem:
(1) The fall of m1 from point A to where it collides with m2
(2) The collision
(3) The rise of m1​

In each step, energy is conserved. In step 2, momentum is also conserved.

Okay, I got it, thanks a lot. When you put it into three steps, something clicked and everything came out. Thanks again.

## 1. What is momentum and why is it important in solving this problem?

Momentum is a physical quantity that measures the motion of an object. In this problem, it is important because it helps us determine the maximum height that m1 will reach after an elastic collision with m2. Momentum is conserved in an elastic collision, meaning that the total momentum of the system before and after the collision remains constant.

## 2. What is an elastic collision and how does it differ from an inelastic collision?

An elastic collision is a type of collision where the total kinetic energy of the system is conserved. This means that after the collision, the objects involved will continue to move without any loss of energy. In contrast, an inelastic collision is a type of collision where the kinetic energy is not conserved. Some of the kinetic energy is converted into other forms of energy, such as heat or sound.

## 3. How do you calculate the maximum height of m1 after an elastic collision?

To calculate the maximum height of m1, we can use the conservation of momentum and energy equations. First, we can solve for the velocity of m1 after the collision using the conservation of momentum equation. Then, we can use this velocity to calculate the maximum height using the conservation of energy equation.

## 4. What are the key assumptions made in solving this problem?

Some key assumptions made in solving this problem include: the collision is elastic, there is no external force acting on the system, and there is no loss of energy due to friction or other factors.

## 5. Are there any real-life applications of this type of problem?

Yes, there are many real-life applications of solving momentum problems, such as in car crashes, sports collisions, and rocket launches. By understanding the principles of momentum, we can predict the behavior of objects in motion and design systems to ensure safety and efficiency.

Replies
8
Views
2K
Replies
7
Views
4K
Replies
8
Views
4K
Replies
2
Views
1K
Replies
4
Views
4K
Replies
17
Views
5K
Replies
1
Views
4K
Replies
10
Views
3K
Replies
5
Views
819
Replies
16
Views
2K