Exploding object (momentum problem)

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SUMMARY

The discussion centers on a physics problem involving the conservation of momentum and kinetic energy related to an exploding projectile. A 5.5 kg stink bomb is launched at an initial speed of 36 m/s at a 30° angle, exploding into two fragments upon reaching its peak. The participants calculate the distance the 3.7 kg fragment lands from the launch point and the energy of the explosion by comparing the kinetic energy before and after the explosion. The correct approach involves using the equations for momentum and kinetic energy, specifically KE=1/2mv², to derive the necessary values.

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TraceBusta
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Well I think conservation of momentum is needed here.

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 5.5 kg projectile launched with an initial speed of 36 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 1.8 kg fragment lands right at the feet of Dave and Jay.

(a) How far from Dave and Jay does the 3.7 kg fragment land?
(b) Find the energy of the explosion by comparing the kinetic energy of the projectiles just before and just after the explosion.

I first solved the x-velocity of the object at the time of explosion (Vx=Vo cos theta) which i think is right. So using that, i solved for P. Then I solved the velocity of the small object and time it took to land at their feet. so i could solve for the velocity of the big object by using 5.5 kg*Vx=3.7 kg*v-(1.8 kg*V1.8) And then I just used that velocity and the same time the other object was in the air to find the distance. I got 99.728m for a but that is wrong. I don't know how to start (b) either.
 
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The time of explosion is when the vertical velocity = 0 (since it is at the top of the path). Use this info when calculating...

part B just involves finding the kinetic energy of the object before it explodes, which is:

KE= 1/2mv^2

THe KE afterwards would just be the same thing, except for both objects.

KE(total) = KE(object 1) + KE(object 2).

Note that the potential energy is not factored into the equation, because the objects before and after the explosion are still at the same height, and thus have the same potential energy, as Potential Energy = mgh.
 


Your approach using conservation of momentum is correct. To solve for the distance, you need to use the same initial velocity and angle to solve for the x and y components of the velocity of the 3.7 kg fragment. Then use those components to find the distance it travels in the x direction. For part (b), you can use the equation for kinetic energy (KE=1/2*m*v^2) to compare the kinetic energy before and after the explosion. The initial kinetic energy would be the sum of the kinetic energy of the 5.5 kg and 1.8 kg fragments, and the final kinetic energy would be the kinetic energy of the 3.7 kg fragment. This will give you the energy of the explosion.
 

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