S.O.S. Mathematics CyberBoard

[Blimpy325]

Can someone please show me what I am doing wrong?


Jay and Dave decide that the best way to protest the opening of a new incinerator is to launch a stink bomb into the middle of the ceremony. They calculate that a 5.2 kg projectile launched with an initial speed of 46 m/s at an angle of 30° will do the trick. The bomb will explode on impact, no one will get hurt, but everyone will stink. Perfect. However, at the top of its flight, the bomb explodes into two fragments, each having a horizontal trajectory. To top it off--this really isn't their day--the 1.9 kg fragment lands right at the feet of Dave and Jay.

How far from Dave and Jay does the 3.3 kg fragment land?
Find the energy of the explosion by comparing the kinetic energy of the projectiles just before and just after the explosion.

I did the following and it was wrong:
h = (46*sin30)^2 / (2*9.81) = 26.9622
Vx = 46*cos30 = 39.8372
Pi(Initial Potential E.) = 5.2Vx = 3.3Vxx - 1.9Vx, Vxx(the velocity of the 3.3kg piece) = 73.6384
T = 46/9.81 = 4.68909
d = (.5T * Vx + .5T * Pi) = 266.0488 meters

[skeeter]

at the top of its trajectory just before the explosion, the entire projectile has a velocity of 46cos(30) m/s.

initial momentum in the x-direction, po = (5.2 kg)[46cos(30) m/s]

immediately upon separation ...

pf = (1.9 kg)[-46cos(30) m/s] + (3.3 kg)(vf)

where both final velocities are in the x-direction.

setting po = pf and solving for vf ...

vf = {(5.2 kg)[46cos(30) m/s] - (1.9 kg)[-46cos(30) m/s]}/(3.3 kg)
vf = approx 85.71 m/s

time to the top of the trajectory is t = 46sin(30)/g = 23/g = approx 2.347 s

the 5.2 kg projectile travels a distance = 46cos(30)*t = approx 93.5 m
the 3.3 kg projectile continues to travel an additional distance = 85.71*t = 201.16m

adding the two gives the distance the 3.3 projectile hits the ground from the position of launch ... approx 294.65 m

at the top of the trajectory, Ko = (1/2)(5.2)[46cos(30)]^2

immediately after separation, Kf = (1/2)(1.9)[46cos(30)]^2 + (1/2)(3.3)(85.71)^2

delta K = approx +9.5 x 10^3 J ... energy added to the system from the explosion

check my calculations.

[Blimpy325]

thanks alot