# Conservation of Momentum Problem- Bullet

## Homework Statement

A 15g bullet strikes and becomes embedded in a 1.10kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is .25, and the impact drives the block a distance of 9.5m before it comes to rest, what was the muzzle speed of the bullet?

## Homework Equations

m1v1+m2v2=m1v'1+m2v'2

## The Attempt at a Solution

I know I will eventually use the above equation in order to solve for V1 but i don't quite understand had to use the other information given such as the friction in order to get to this point in the problem solving process. Any advice please?

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Did you try using newtons second law, F=ma to find the acceleration in the x direction?

You know the coefficient of friction and you can find the normal force of the block and bullet combo to get the Fn. So Fn(coefficient of friction)= Force friction.

Then use kinematic eq's to get the blocks velocity, then use momentum.

You use that equation to calculate the instantaneous velocity of the bodies right after the impact. After this friction is there, so there will be acceleration. So you need the friction to calculate the acceleration of the body, and since the distance of stoppage is given, you can calculate the initial velocity, from wich using the equation you described, the speed of the bullet can be calculate before the impact.

Did you try using newtons second law, F=ma to find the acceleration in the x direction?

You know the coefficient of friction and you can find the normal force of the block and bullet combo to get the Fn. So Fn(coefficient of friction)= Force friction.

Then use kinematic eq's to get the blocks velocity, then use momentum.

Sorry but I'm still a bit confused. I know that Newton's Second Law is F=ma, but how can you find the normal for of the combo. Could you explain this a bit further?

Sorry but I'm still a bit confused. I know that Newton's Second Law is F=ma, but how can you find the normal for of the combo. Could you explain this a bit further?
Sure.

You know that the bullet is embedded in the block so they have a certain combined mass. So you are given the coefficient between the surface and the block and you know that Ffr=Fn(coefficient friction)... So when the bullet goes into the block, it accelerates forward, but forces in the Y axis are still balanced...

So Fw=Fn, and then from that you can find the Force of Friction and use F=ma in the X direction to find the acceleration.

Then you can use that accel to find the initial speed of the block bullet system...

and then use mv(1)+mv(2)=V(m1+m2) to get the initial v1 because you already will have solved for the V when you found the initial speed of the bullet block system from the kinematic equation.

LowlyPion
Homework Helper
A 15g bullet strikes and becomes embedded in a 1.10kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is .25, and the impact drives the block a distance of 9.5m before it comes to rest, what was the muzzle speed of the bullet?

## Homework Equations

m1v1+m2v2=m1v'1+m2v'2

## The Attempt at a Solution

I know I will eventually use the above equation in order to solve for V1 but i don't quite understand had to use the other information given such as the friction in order to get to this point in the problem solving process. Any advice please?
First solve the retarding force (F = ma = μmg ) The acceleration (deceleration in your case) then can be used with standard kinematic equation like Vf² = Vi² + 2*a*x

This yields your initial velocity after the bullet strikes the block. From that use the conservation of momentum to figure the initial speed.