# Block being shot upwards

1. Oct 4, 2007

### turnip

1. The problem statement, all variables and given/known data
question: a 2kg block rests over a small hole in a table. janelle is breathe the table and shoots a 15g bullet through the hole and into the block, where it lodges. How fast was the bullet going if the block rises 75cm above the table?

2. Relevant equations
m1v1i +m1v2i = m1v1f + m2v2f

3. The attempt at a solution
i tired working this out with the straight line motion equation of s= vt -at^{2}
but this did not workout

i tired solving for t but got .153s
from there i am lost

2. Oct 5, 2007

### learningphysics

There are 2 parts to the problem...

1) bullet being fired into block... this requires conservation of momentum.

2) bullet+block rising up to 75cm... this requires conservation of energy.

try to write the equations for part 1) and part 2)... then you should be able to solve for the initial velocity of the bullet.

3. Oct 5, 2007

### turnip

okay so m1v1i = m1v1f + m2v2f
but (because they are combined and therefor have the same combined mass and velocity), couldn't it be written as: m1v1i = (m1 + m2) vf

the conservation of energy? as in Ke=Ep or 1/2 mv(squared) = mgh?
or as in initial Ke= final Ke?

there are two unknows here. i guess what your saying is work them (the the separate equations) out and substitute it for one unknown. but could you please give me a bit of guidance? one i see it once i'll be good from there on.

4. Oct 5, 2007

### Staff: Mentor

Last edited: Oct 5, 2007
5. Oct 5, 2007

### learningphysics

you're almost there:

equation 1:

m1v1i = (m1 + m2)vf

equation 2:

(1/2)(m1 + m2)vf^2 = (m1 + m2)gh (as Astronuc mentioned, all the energy is converted to gravitational potential energy)

solve these two equations... you can cancel the masses in the second equation. your two unknowns are v1i and vf...

6. Oct 6, 2007

### turnip

right, that makes a lot of sense to me now
just rearranged for vf and subs into v1i and i get the right answer
thanks to both of you