Conservation of momentum - bullet/pendulum

1. Jun 15, 2011

z2420

1. The problem statement, all variables and given/known data
Bullet = 0.02 kg
pendulum block = 2 kg

bullet fired from (0m,0m) with velocity of (x m/s, 30 deg)
pendulum block located (30m, 30deg)
the maximum height reached by pendulum is 0.2 meter

what is the value of X

2. Relevant equations
I've tried several ways to approach this problem. the problem is the angle that it hits the bullet.

3. The attempt at a solution

If the angle was out of the problem, i get an answer of 199.98 m/s/s. i used mgh=1/2mv^2 to get the velocity final, then worked backwards to plug that into the conservation of momentum equation:

(0.02)(Vi) = (2.02)(1.98) = 199.98

this is about as far as i got without running into a wall. What i think is that the 199.98 is the X component, and using that I can solve for the resultant. But how?

2. Jun 15, 2011

Staff: Mentor

Hi z2420, Welcome to Physics Forums.

It's not clear how this pendulum is supposed to look. How long is the pendulum string? Where does it pivot? Where is its "maximum height" measured from?

If the pendulum block is represented as a point mass at location 30m @30°, and the bullet is fired from the origin at 30°, how can the bullet ever strike the block if gravity is acting? (Trajectory will be parabolic, always below a slope 30° line from the origin). Is the bullet striking the block an elastic or inelastic collision?

If the pendulum string is just 0.1 meters long, any sufficiently speedy bullet will always make it reach its apex of 0.2m above its lowest point; The pendulum will go around in circles about its pivot!

Did the original problem statement come with a diagram?

3. Jun 15, 2011

z2420

there was no diagram and the problem stated was very unclear in itself

this is what i thought

the bullet in traveling upward at an angle of 30 degrees. Hitting the block and going inside. The block and bullet swing upwards at 0.2m. The length of the pendulum string is not stated. the pendulum looks like a string hanging 90 degrees from the top.

i think the 199.98 is the x trajectory. So under this (maybe incorrect) assumption, i calculated the y trajectory to get the resultant = 230.91 = initial velocity of bullet

think thats right?

4. Jun 15, 2011

Staff: Mentor

I think that there are too many inconsistencies and unknown properties in the problem to make a valid judgment. Sorry.