# How Does Holding an Air Rifle Affect Recoil Speed?

• dlilpyro
In summary, the conversation discusses the calculation of the recoil speed of a 2.0 kg air rifle when fired with a bullet of mass 1.0 g at a muzzle velocity of 150 m/s. The second part involves holding the rifle tightly against the body and calculating the new recoil speed, taking into account the mass of the girl (48 kg) as part of the system. Momentum is a key factor in solving this problem and the equation is set up to account for the momentum of the bullet and the girl moving backwards with some velocity. The initial velocity of the girl is considered to be zero, but there will be a different final velocity for the girl due to the recoil.
dlilpyro
Kim holds a 2.0 kg air rifle loosely and fires a bullet of mass 1.0 g. The muzzle velocity of the bullet is 150 m/s. What is the recoil spped of the rifle? The next shot the girl holds the rifle tightly against her body. Calculate the new recoil speed assuming the girl has a mass of 48kg.

I'm really not sure how to do this problem, any help would be appreciated.
Thanks

Well you know that momentum will play a part in this question, so will momentum be conserved as the bullet leaves the gun?

Yes so is this the right way to set up the equation for the second part:
(50)(v)=(.001)(150)
v=-.003 m/s - v is negative because moving backwards

I'm not sure if 50 kg is the right mass though. I took it as the girl and the gun becoming one system so their masses are added.

dlilpyro said:
Yes so is this the right way to set up the equation for the second part:
(50)(v)=(.001)(150)
v=-.003 m/s - v is negative because moving backwards

I'm not sure if 50 kg is the right mass though. I took it as the girl and the gun becoming one system so their masses are added.

Her initial mass would be 48+2+0.001 since she is holding the gun with the bullet in it. However, what is her initial velocity if she is not moving?

After the bullet is fired, you are right with the 0.001(150) but remember she will move backwards with some velocity. So you need to have an additional term on the right side.

I'm not sure what the additional term is because we didn't have time to get the equation during class today. But if the initial velocity is zero how can you solve for v final on the left side?

dlilpyro said:
I'm not sure what the additional term is because we didn't have time to get the equation during class today. But if the initial velocity is zero how can you solve for v final on the left side?

The v on the left side will be zero since the girl is not moving.

There will be a different V on the right side since the bullet will move forward and she will move backward due to the recoil.

on the right side you will need the moment of bullet (which you have already) + momentum of girl (which will have the V to solve).

Alright I think I have it now. Thanks for all the help.

## 1. What is momentum?

Momentum is a measure of an object's motion, and it is equal to the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

## 2. How is momentum calculated?

Momentum (p) is calculated by multiplying an object's mass (m) by its velocity (v), using the formula p = m * v.

## 3. What is impulse?

Impulse is the change in an object's momentum over a period of time. It is equal to the force applied to an object multiplied by the time it is applied for, represented by the formula J = F * Δt.

## 4. How does impulse relate to momentum?

Impulse and momentum are closely related. In fact, impulse is defined as the change in momentum, or the product of force and time. This means that the greater the impulse on an object, the greater its change in momentum will be.

## 5. What are some real-world applications of momentum and impulse?

Momentum and impulse are important concepts in physics and have many real-world applications. They are used in sports, such as calculating the force needed to kick a soccer ball into the goal, and in car safety, as cars are designed with crumple zones to increase the time of impact and reduce the impulse on passengers during a collision.

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