1. Nov 29, 2003

### formulajoe

a projectile of mass 3 kg is fired at 120 m/s at an angle of 30 deg. at the top of the trajectory the projectile explodes into 1 kg and 2 kg fragments. the 2 kg fragment lands directly below the point of explosion. it takes 3.6 sec for the 2 kg object to reach earth. find the distance between the point of firing and the point at which the 1 kg object strikes the ground.

Ive figured out where the 2 kg object lands to be 634 meters from the point of firing. The top of the trajectory is 184 meters up. When the object explodes the 2 kg object takes some of the KE because it only takes 3.6 sec to reach the ground. So using conservation of energy i set up an equation like this
1/2(3kg)(v^2) = 1/2(2kg)(33.5^2) + 1/2(1 kg ) v1^2.
I'm trying to find v1 so i can find the distance. but i dont know what to use for v. do i use the 120 m/s or do i use the horizontal component of the velocity which is 104 m/s?

2. Nov 29, 2003

### jamesrc

Use the conservation of momentum, realizing that the velocity of the projectile at the point of the explosion is purely in the x direction. The momentum of the 2kg piece after the explosiotn is purely in the y-direction. The sum of this momentum and the y-momentum of the 1-kg piece must be 0. The x-component of the momentum of the 1 kg piece is equal to the original x-momentum of the unexploded projectile. With the velocity components of the 1 kg piece and its initial position, the rest of the problem is just a projectile motion exercise.

3. Nov 29, 2003

### formulajoe

using conservation of momentum i get the velocity of the 1 kg piece to be 247 m/s.
i did (3kg)(104m/s) = (2kg)(33.5) + 1kg(v).
the 104 is the x component of the velocity. the 33.5 is the y component of the 2 kg piece. this leaves the velocity of the 1 kg piece 247. is that right? that sounds too high.

4. Nov 29, 2003

### jamesrc

Remember to treat momentum as a vector quantitiy:

Just before the explosion, the momentum of the projectile is 3kg*120m/s*cos(30 degrees) in the x direction. Since the 2 kg piece goes straight down, conservation of momentum in the x direction gives: 3*120*cos(30) = 2*0 + 1*v1,x. Solve that to get the x-velocity of the 1 kg piece.

For the y-momentum, just before the explosion v3,y = 0. v2,y comes from the kinematic data given (you already found this velocity to be about -33 m/s (note the negative sign, implying the initial velocity is down). Conservation of momentum now gives: 0 = 2*v2,y + 1*v1,y. Solve that to get the y-velocity of the 1 kg piece.

Then use that info to find the landing point.