Calculating Acceleration in a Rifle-Bullet System

[Justhelp]

Homework Statement

A rifle with a mass of 5.0kg fires a bullet with the mass of 2.0g.The bullet has an acceleration of 2900m/s^2. What is the acceleration of the rifle?

Homework Equations

I am pretty sure I have honed in on the formula to use, I am not sure how to use it though.

Also, is acceleration force?

The Attempt at a Solution

M=5.0kg
M=2.0g
A or F=2900

a=f/m or f/m=a

fyi: /sign is over.

Thank you

 

[berkeman]

Homework Statement

A rifle with a mass of 5.0kg fires a bullet with the mass of 2.0kg.The bullet has an acceleration of 2900m/s^2. What is the acceleration of the rifle?

Homework Equations

I am pretty sure I have honed in on the formula to use, I am not sure how to use it though.

Also, is acceleration force?

The Attempt at a Solution

M=5.0kg
M=2.0KG
A or F=2900

a=f/m or f/m=a

fyi: /sign is over.

Thank you

Force is not acceleration. They are related by the equation F=ma.

And I'm pretty sure that you have a typo for the mass of the bullet. Morel likely it is 2g = 0.002kg, right? Otherwise, that is one honkin' big bullet for that gun!

 

[Justhelp]

F=MA if that is it and seems more like it then it would work out to
F= 5.0kg(2900m/s^2)

=14500...but it's not asking for the force but the acceleration of the rifle... so the formula is switched around a bit. hm...

 

[Justhelp]

Is it bullet mass x bullet acceleration? That would be 5800 the force of the bullet right?
Then the formula moves around to become a=f(m), a=5800(5.0) a=29000 now.

 

[Justhelp]

And.. the mass of the bullet is 2.0g. Whoops.

 

[berkeman]

F=MA if that is it and seems more like it then it would work out to
F= 5.0kg(2900m/s^2)

=14500...but it's not asking for the force but the acceleration of the rifle... so the formula is switched around a bit. hm...

You are close. They are assuming that the rifle is not supported by a shooter or anything as the bullet is fired. They want you to equate the forces of the rifle on the bullet and the bullet on the rifle. So you are given the mass and acceleration of the bullet -- what does that give you for the force of the rifle on the bullet?

Then use that force as the force of the bullet on the rifle -- you know the mass of the rifle, so what does that give for its acceleration?

They are trying to illustrate how given the same force, the acceleration depends on the mass...

 

[Justhelp]

"Then the formula moves around to become a=f(m), a=5800(5.0) a=29000 now."

29000m/s^2 right?

 

[berkeman]

"Then the formula moves around to become a=f(m), a=5800(5.0) a=29000 now."

29000m/s^2 right?

I don't understand what you posted. What is the force that you get from the bullet's acceleration and mass?

 

[Justhelp]

f=(2g)(2900m/s^2) = 5800 the force of the bullet, then I did this accel= fm

a=5800(5.0)

 

[berkeman]

f=(2g)(2900m/s^2) = 5800 the force of the bullet, then I did this accel= fm

Check your units. You cannot multiply grams and m/s^2 -- you need to keep the masses in kg...

(and label the answer with the units of Newtons 1N = 1kgm/s^2)

When you get the force answer, calculate the acceleration of the rifle...

 

[Justhelp]

the mass is grams... I'm confused. If I can't multiply m/s^2 with gram then what can I do? Do I use the 5.0kg instead?

 

[berkeman]

the mass is grams... I'm confused. If I can't multiply m/s^2 with gram then what can I do? Do I use the 5.0kg instead?

You need to work in a consistent set of units. The SI units that you will typically use for intro physics problems is meters, kilograms, seconds (mks). You need to keep all of your masses in kilograms, in order to be able to get the right answers.

So if the bullet's mass is 2g, what is that mass in kg? And proceed...

 

[Justhelp]

F=MA
F= 5.0kg(2900m/s^2)

=14500 is the force of the rifle? I have to still use that number? Earlier I had assumed you meant that that was wrong.

 

[Justhelp]

Does this mean the 2.0g has to be translated into Kg?

 

[Justhelp]

and what I had had was correct? heh... but just not the unit.

 

[berkeman]

F=MA
F= 5.0kg(2900m/s^2)

=14500 is the force of the rifle? I have to still use that number? Earlier I had assumed you meant that that was wrong.

That is not correct or logical. You are mixing the mass of the rifle and the acceleration of the bullet. They have nothing to do with each other...

Does this mean the 2.0g has to be translated into Kg?

Yes.

and what I had had was correct? heh... but just not the unit.

What you had (see below) was correct for the force of the rifle on the bullet, if you convert the 2g mass into its equivalent mass in kg...

f=(2g)(2900m/s^2) = 5800 the force of the bullet,

 

[Justhelp]

AH! ok, I didn't know that m/s^2 could not be multiplied by g, only kg mass. Sorry for any frustration.