# Conservation of Energy bullet of mass

1. Mar 19, 2007

### chaoseverlasting

1. The problem statement, all variables and given/known data
A bullet of mass m is fired with a velocity of 50m/s at an angle x. At the top of its trajectory it hits a bob of mass 3m hanging by a string 3.3 m long and gets imbedded in it. The string rotates by an angle of 120 degrees.

Find x and the coordinates of the bob wrt the projectile.

3. The attempt at a solution

The bullet hits the bob with velocite 50 cosx. The kinetic energy of the bullet gets converted into the gravitational potential energy of the bob and the bullet.

$$0.5m(50cosx)^2= 4mg*3.3*(1+sin30)$$
cosx=0.4

The answer, however, is 30 degrees. I used cosx=0.4 to solve the second part, but needless to say, that answer didnt match either.

What did I do wrong? Or is the given answer incorrect?

2. Mar 19, 2007

### Mindscrape

I have no idea what "bob wrt the projectile" means, and so I am not really sure what you did wrong or what you were trying to solve. However, in your energy equation it looks like you are doing
$$E_i = E_f$$
$$K_i + U_i = K_f + U_f$$
$$K_i = U_f$$

I agree with the initial kinetic, but it looks to me like you calculated the height the wrong for your potential.

3. Mar 19, 2007

### Staff: Mentor

Good.
Careful! Is energy conserved during the collision of bullet with bob?

4. Mar 20, 2007

### chaoseverlasting

Isnt the energy conserved? And bob wrt projectile means that the origin is at the point where the projectile was launched. We have to find the position of the bob w.r.t that origin.

5. Mar 20, 2007

### e(ho0n3

Only momentum is conserved, not energy.

6. Mar 20, 2007

### Staff: Mentor

As e(ho0n3 has stated, during the collision only momentum is conserved. But after the collision, the mechanical energy of the "bullet+bob" system is conserved. Both parts are needed to solve this problem.