Conservation of momentum problem (I think?)

[downwithsocks]

Homework Statement

http://img14.imageshack.us/img14/5489/25765454.png

A projectile is launched from a cliff above level ground. At launch the projectile is 35 meters above the base of the cliff and has a velocity of 50 meters per second at an angle 37° with the horizontal. Air resistance is negligible. Consider the following two cases and use g = 10m/s², sin37° = 0.60, and cos37° = 0.80.

(Case I includes parts a. b. and c., which I don't need help on so will skip.)

Case II: A small internal charge explodes at point B in the following diagram, causing the projectile to separate into two parts of masses 6kg and 10kg. The explosive force on each part is horizontal and in the plane of trajectory. The 6kg mass strikes the ground at point D, located 30 meters beyond point C, where the projectile would have landed had it not exploded. The 10kg mass strikes the ground at point E.

http://img194.imageshack.us/img194/2348/23075084.png

d. Calculate the distance x from C to E.

Homework Equations

Assuming it is conservation of momentum,
KE[tex]_{i}[/tex] = KE[tex]_{f}[/tex]
KE = (1/2)mv²
m1v1[tex]_{i}[/tex] + m2v2[tex]_{i}[/tex] = m1v1[tex]_{f}[/tex] + m2v2[tex]_{f}[/tex]

The Attempt at a Solution

In part a., I've calculated that the total time from launch to landing at point C is 7s. In b., that the distance R from launch to C is 280m. And in c., V[tex]_{A}[/tex] = 40m/s, V[tex]_{B}[/tex] = 50m/s, and V[tex]_{C}[/tex] = 56.6m/s. All of these parts just required simple kinematics and logic.

But for part d., I'm stuck. I tried conservation of energy, using the velocity at B and the given masses to try to find the velocities of each of the particles after the explosion, but doing this I somehow proved that both velocities remain 50m/s (I think this means that my calculation assumed the explosion did not cause the particles to separate). I'm not sure where to go.

 

[rl.bhat]

At B the velocity of the projectile is the same as the launched velocity and the angle is 37 degrees below horizontal. It explodes. Let V2 be the velocity of the 6 kg mass in the original direction. Find range up to B. Range up to C is known. Find BD.
For 6 kg mass, angle of projection, range and y is known. Find V2.
Using equation of conservation law, find velocity v1of 10 kg mass.. Then proceed in the same way as above to find range of 10 kg mass.

 

[troy611]

I think to solve this problem, u need to know center of mass concept

Now, as the explosive force on each part is horizontal and in the plane of trajectory (it simply means that no external force is acting on the system, except the gravitational force). Now since no external force is acting, except the gravitational force, the center of mass of the two broken masses continues on the original path(the path which the unbroken mass would have followed).

So, the two broken masses land in such a way that their center of mass is 280 m from the point of projection.

To simplify calculations, I have taken center of mass co-ordinate as 0, so co-ordinate of one mass is (30, 0), and other is (-x, 0)

Using center of mass definition:

0 = [(6*30) + (10*-x)] / 16

solving, x=18

So, the other mass lands 18 m away from the point C.

 

[downwithsocks]

Ah, I didn't even think to use center of mass...I should probably review a little, but at least it completely makes sense now, thanks :)

 

[troy611]

anytime...:)