# Trouble understanding collisions

• Opus_723
In summary, the individual was initially confused about a lab question regarding the conservation of momentum and kinetic energy in a collision between a light mass and a heavy mass. After further investigation, they realized that the situation presented in the lab was unphysical and they needed to approach the problem in more general terms. They found that the light mass would have a velocity of -kv after the collision, and in order for energy to be conserved, k must be equal to (M-m)/(M+m). This value can never equal one, but approaches one as the difference between the masses increases. Ultimately, the individual was able to solve the problem and understand the concept.

#### Opus_723

So we had some questions in lab today that left me confused. Someone please tell me what I'm not understanding here?

We start with a light mass moving at velocity v, having linear momentum p=mv and kinetic energy (mv^2)/2, and a heavy mass at rest, with linear momentum p=M(0)= 0 and kinetic energy (m(0^2))/2= 0. The total momentum of the system is 0 + mv = mv, and the total kinetic energy is 0 + (mv^2)/2 = (mv^2)/2.

Suppose the light mass collides with the heavy mass in such a way that it has a new velocity nearly equal to its original velocity but opposite in direction. It's new momentum is m(-v)= -mv, and its new kinetic energy is (m(-v)^2)/2= (mv^2), which is the same kinetic energy it began with. Since the initial momentum of the system was p = mv, we have
mv = x -mv, where x is the final momentum of the heavy mass. So x = 2mv. Since the heavy mass has a momentum, it must have kinetic energy. But the light mass has the same kinetic energy that it had at the start, so adding any energy to that would give us more kinetic energy than we started with.

What did I miss?

Maybe I spoke too soon. I think I figured it out myself.

My first suspicion was that the situation the lab gave me was unphysical. That an object could never bounce back with the same speed it started with.

So I decided to work out the problem in more general terms so that if you start with the situation above, the light mass has a velocity -kv after the collision for some value of k. Then I applied conservation of kinetic energy and found that in order for energy to be conserved, k must be equal to (M-m)/(M+m). This is zero for equal masses and approaches one as the difference between the masses increases, but can never equal one.

So yay!

EDIT: For anyone wondering why I posted for help when I had an idea of where to go next, it's because the first time I tried the above, I lost a negative somewhere and got a silly result that suggested that objects ALWAYS bounce back with equal speed. After posting, I decided to double check my work. I'll just let that be a lesson to myself.

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## 1. What is a collision?

A collision occurs when two or more objects come into contact with each other and exert forces on each other, resulting in a change in their motion.

## 2. What are the different types of collisions?

There are two main types of collisions: elastic and inelastic. In an elastic collision, the total kinetic energy of the objects before and after the collision remains the same. In an inelastic collision, some kinetic energy is lost and the objects may stick together.

## 3. How do you calculate the momentum of a collision?

The momentum of a collision is calculated by multiplying the mass of an object by its velocity. In a multi-object collision, the total momentum before the collision is equal to the total momentum after the collision (conservation of momentum).

## 4. What factors affect the outcome of a collision?

The outcome of a collision can be affected by several factors, including the mass, velocity, and direction of the objects involved, as well as the type of collision (elastic or inelastic).

## 5. How can collisions be used in real-life applications?

Collisions are used in many real-life applications, such as in car safety features, sports equipment design, and even in particle accelerators used in scientific research. Understanding collisions can also help us predict and prevent potential accidents or collisions in various scenarios.