# Conservation of momentum 2 step problem

1. Dec 14, 2008

### Maiia

1. The problem statement, all variables and given/known data
A 0.0854 kg block is released from rest from the top of a 25.8◦ frictionless incline. When
it has fallen a vertical distance of 0.905 m, a 0.0164 kg bullet is fired into the block along a
path parallel to the slope of the incline, and momentarily brings the block to rest, stopping in the block. The acceleration of gravity is 9.8 m/s2. Find the speed of the bullet just before impact. Answer in units of m/s.

I was hoping someone could check my work because I seem to be getting the wrong answer...

What I did:
1. found the height of the incline
sin25.8=x/.905m
x=.3938841449m

mgh=1/2mv^2
$$\sqrt{2gh}$$=v
v= 2.778512055m/s

2. used conservation of momentum
m1= block
m2=bullet
After having fallen the .905m
m1v1initial+m2v2initial= m1+m2(Vf)
because Vf=0
m1v1initial= m2v2initial
so plugging in m1= .0854kg m2=.0164kg v1=2.778512055m/s
i got v2= 14.46859326m/s

2. Dec 14, 2008

### LowlyPion

Haven't they conveniently already provided you with the drop in height? Won't kinetic energy of the block be the .905*m*g then? And your V2 = 2*g*.905 ?

3. Dec 14, 2008

### Maiia

oh..i didn't read the problem carefully enough and assumed they were giving me the amount it was falling on the slope. Thanks for your help :)

4. Dec 14, 2008

### Maiia

oh, if i wanted to find the speed of the bullet that would push the block all the way back up the incline back to its initial position, would i be able to do this? :
1/2m1v^2= m2gh
mass1 i am denoting as mass of bullet
m2 i am denoting as combined mass
so vbullet= squareroot of (2*.1018*9.8*.905/ .0164) ?
my reasoning being that the ke of bullet is converted to pe needed to push block back up incline?

5. Dec 14, 2008

### LowlyPion

If you mean additional amount of energy (more than just stopping it), you should also consider that you need to add the mass of the bullet to the total mass pushed up the incline.