# Simple momentum problem

• crybllrd
In summary, a small block with a mass of 4kg and a large block with a mass of 12kg collide head on a frictionless table. Before the collision, the small block is traveling 8m/s to the right and the large block is traveling 4m/s to the left. If the small block bounces back at 3m/s to the left, the final velocity of the large block can be calculated using the equation p=mv, resulting in a final velocity of -1m/s. If the two blocks stick together after the collision, the total energy lost can be calculated using the equation KE=\frac{1}{2}mv^2, resulting in a loss of 216J.

## Homework Statement

A small block m=4kg and a large block m=12kg have a head on collision on a frictionless table. Before the collision, the small block is traveling 8m/s to the right and the large block is traveling 4m/s to the left.

A. If the small block bounces back at 3m/s to the left, find the final velocity of the large block.

B. Suppose instead the two blocks stick together after they collide. How many Joules of energy would be lost?

## Homework Equations

$$p=mv$$

$$KE=\frac{1}{2}mv^{2}$$

## The Attempt at a Solution

I will define to the right to be the positive x direction.
First I will calculate initial momentum for both blocks.

$$p_{i a}=mv$$
$$p_{i a}=(4kg)(8m/s)$$
$$p_{i a}=32kg m/s$$

$$p_{i b}=mv$$
$$p_{i b}=(12kg)(-4m/s)$$
$$p_{i b}=-48kg m/s$$

$$p_{i}=-16kg m/s$$

Now final momentum:

$$p_{f a}=mv$$
$$p_{f a}=(4kg)(-3m/s)$$
$$p_{f a}=-12kg m/s$$

$$p_{f b}=mv$$
$$p_{f b}=(12kg)v$$
$$p_{f b}=12v$$

$$p_{f}=-12kg m/s + 12v$$

If momentum is conserved, then

$$p_{i}=p_{f}$$

$$-16kg m/s=-12kg m/s + 12v$$

$$v=\frac{-1}{3}$$

So far you are correct. Is there a question you wish to ask?

Oh, great. I had a feeling I was on the wrong track.

For part B, I need to find the energy. It is all Kinetic.$$KE_{i}=\frac{1}{2}mv_{a}^{2}+\frac{1}{2}mv_{b}^{2}$$

$$KE_{i}=\frac{1}{2}(4)(8)^{2}+\frac{1}{2}(12)(4)^{2}$$

$$KE_{i}=128+96$$

$$KE_{i}=224J$$

Now the final kinetic energy is where I am getting confused.

$$KE_{f}=\frac{1}{2}mv_{a}^{2}+\frac{1}{2}mv_{b}^{2}$$

How can I find the velocity of the two blocks?

Actually, for part B I did it another way:

$$p_{i}=32+-48$$

$$p_{i}=-16$$

$$p_{f}=16v$$

$$p_{i}=p_{f}$$

$$-16=16v$$

$$v=-1m/s, or 1 m/s to the left$$

$$KE_{i}=\frac{1}{2}mv^{2}+\frac{1}{2}mv^{2}$$

$$KE_{i}=\frac{1}{2}(4)(8)^{2}+\frac{1}{2}(12)(4)^{2}$$

$$KE_{i}=224J$$

$$KE_{f}=\frac{1}{2}mv^{2}$$

$$KE_{f}=\frac{1}{2}(16)(1)^{2}$$

$$KE_{f}=8J$$

$$KE_{i}-KE_{f}=E_{lost}$$

$$KE_{i}-KE_{f}=E_{lost}$$

$$224-8=E_{lost}$$

$$E_{lost}=216J$$

## 1. What is a simple momentum problem?

A simple momentum problem involves calculating the momentum of an object using its mass and velocity. Momentum is a measure of an object's motion and is equal to its mass multiplied by its velocity.

## 2. How do I solve a simple momentum problem?

To solve a simple momentum problem, you will need to use the formula p = mv, where p is momentum, m is mass, and v is velocity. Simply plug in the given values and solve for the missing variable.

## 3. Can you give an example of a simple momentum problem?

Sure, for example, if a car with a mass of 1000 kg is traveling at a velocity of 20 m/s, what is its momentum? Using the formula p = mv, we can calculate the momentum to be 20,000 kg m/s.

## 4. How is momentum related to Newton's laws of motion?

Momentum is directly related to Newton's laws of motion, specifically the second law which states that the net force on an object is equal to the rate of change of its momentum. This means that an object with a larger momentum will require a greater force to stop or change its motion.

## 5. How is momentum conserved in a simple momentum problem?

In a simple momentum problem, momentum is conserved, meaning that the total momentum of all objects in the system remains constant. This is known as the law of conservation of momentum and is a fundamental principle in physics.