- #1

TheLegace

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## Homework Statement

The muzzle velocity of a gun found at a crime scene is tested by firing the 100 g bullet

into a block of mass 9.9 kg. The block is initially at rest on a frictional surface,

coefficient of 0.1. The bullet sticks into the block and the combination slides a total

distance of 4.5 metres along the surface. Calculate the muzzle velocity of the gun.

## Homework Equations

This collision is inelastic.

Fy = 0 = Fn = Fg = mg(Force in y)

Ff = uFn = (.1)(9.9kg)(9.8m/s^2) = 9.703N (Friction Force)

Fx = ma = (-Ff) + (Fa)

vx = dx/t

mBvB + mLvL = v'(mB + mL) (B - bullet, L - block)

mBvB = v'(mB + mL) (vL = 0)

## The Attempt at a Solution

I know the frictional force, if I could find a way to figure out the v' in the momentum equation, I could figure everything out. Now I know the displacement, if there was some way I could find the time, or even acceleration. I think I might have ideas on how to taclke the problems, but I can't get too far, I am still going to try, if anyone could help that would very well appreciated. Thank You.

I am missing a piece of the puzzle to finish the question.

## Homework Statement

Two titanium spheres approach each other head-on with the same speed and collide

elastically. After the collision, one of the spheres, whose mass is 300 g, remains at rest.

What is the mass of the other sphere?

## Homework Equations

Momentum Conserved

m1v1 + m2v2 = m1v1' + m2v2' (v2' = 0, v1 = v2, m1 = .3kg)

v(m1+m2) = m2v2'

m1v1^2 + m2v^2 = m2v2'^2(should I factor?)

v^2(m1 + m2) = m2v2'^2

## The Attempt at a Solution

Right now I have a feeling that I should be isolating for m2 or something so I can cancel it out when I sub into another equation.

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