# Momentum: Bullet and Block

#### huynhtn2

1. The problem statement, all variables and given/known data
When a bullet strikes a block of wood originally at rest, the bullet becomes embedded into the wood block which travels 1.70 m after the collision. If the mass of the bullet is 5.26 g and the mass of the block is 2.40 kg, what is the speed of the block immediately after the collision? The coefficient of kinetic friction between the block and the surface is 0.30.

2. Relevant equations
p=mv
f=dp/dt
m1v1 + m2 v2 = m1u1 + m2u2

3. The attempt at a solution

m1v1 + m2 v2 = m1u1 + m2u2
(0.00526)v1 = (0.00536)u1 + (2.4)u2

#### PhanthomJay

Homework Helper
Gold Member
This is a totally inelastic collision where the block and bullet move, after the collision , together as one mass with a common speed. Then you'll need another equation or two to solve for that speed.

#### huynhtn2

This is a totally inelastic collision where the block and bullet move, after the collision , together as one mass with a common speed. Then you'll need another equation or two to solve for that speed.
So i got mvi = (M + m)vf
and 0.5mv^2=kfriction
vf= 2.43 m/s
then i plugged that in to the first equation for vf and solved for vi, is this correctÉ

#### Xerxes1986

You can't use conservation of momentum because they didn't give you the initial OR the final velocities in the problem. Sure it would work if you had that information, but because you don't you have to approach the problem a little differently. Think about it from an energy perspective. Right after the collision the block and bullet have some velocity $$v_{i}$$ and therefore some Kinetic Energy $$\frac{1}{2}mv^{2}_{i}$$. Then the block travels some distance on the table to a final position 1.70m from start and stops. This means that you have lost all that kinetic energy. It had to go somewhere.......(hint, Work due to friction)

#### Parbat

You can't use conservation of momentum because they didn't give you the initial OR the final velocities in the problem. Sure it would work if you had that information, but because you don't you have to approach the problem a little differently. Think about it from an energy perspective. Right after the collision the block and bullet have some velocity $$v_{i}$$ and therefore some Kinetic Energy $$\frac{1}{2}mv^{2}_{i}$$. Then the block travels some distance on the table to a final position 1.70m from start and stops. This means that you have lost all that kinetic energy. It had to go somewhere.......(hint, Work due to friction)
u r right.the energy in the bullet is used to take the block & the bullet 1.70 m ahead.
so u can equate the kinetic energy of bullet & work done by "bullet & block" to travel 1.70 m against friction.
use: Kfriction=coefficient of friction * weight of (block+bullet)
& 0.5 mv^2=Kfriction*1.70

#### huynhtn2

u r right.the energy in the bullet is used to take the block & the bullet 1.70 m ahead.
so u can equate the kinetic energy of bullet & work done by "bullet & block" to travel 1.70 m against friction.
use: Kfriction=coefficient of friction * weight of (block+bullet)
& 0.5 mv^2=Kfriction*1.70
Ke = Ffriction
0.5mv^2= uk mg
v= square root (2(9.81)(0.3))
v= 2.43 m/s

i dont know what to do with the 1.70 m distance

#### Parbat

Ke = Ffriction
0.5mv^2= uk mg
v= square root (2(9.81)(0.3))
v= 2.43 m/s

i dont know what to do with the 1.70 m distance
that's there:
0.5mv^2=Kfriction*1.70
u see that?

#### ghostanime2001

"When a bullet strikes a block of wood originally at rest, the bullet becomes embedded into the wood block which travels 1.70 m after the collision."

Do they mean the bullet + wood block system move 1.70 m after the collision ? OR the bullet travels 1.70 m inside the wood block and the bullet and wood block move together at the same time as the bullet is getting further embedded into the wood block ?

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