Momentum of cannon ball homework

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A cannonball explodes into two pieces, one weighing 3 lb and the other 5 lb, after reaching a height of 875 feet. The 5 lb piece lands 1,659 feet away from the explosion, and the work done before the explosion is calculated to be 224,000 J. The work done after the explosion is divided between the two pieces, with the 5 lb piece accounting for 140,000 J. The discussion emphasizes using momentum principles and projectile motion equations to solve for the velocity of the 5 lb piece post-explosion. The problem can be approached through both momentum conservation and energy principles, though momentum is highlighted as the key focus.
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A cannon shoots a shell straight up. It reaches its maximum height, 875 feet, and splits into two pieces, one weighing 3 lb and the other 5 lb. The two pieces are observed to strike the ground simultaneously. The 5 lb piece hits the ground 1,659 feet away from the explosion (measured along the x axis). Find the magnitude of the velocity of the 5 lb piece just after the explosion.

Work done before explosion = 8lb * 32 ft/s^2 * 875 ft = 224000 J

work done before explosion = work done after explosion

work done after explosion = work done by 3lb + work done by 5 lb = 224000 J

work done by 3lb= 3lb * 32 ft/s^2 * 875 ft = 84000 J

work done by 5 lb = 224000 - 84000 = 140000 J

work = integral of force done
work done by 5 lb = integral of force in x direction + integral of force done in y direction

where do I go from here?
 
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Convervation of energy is not required for this question. Have you met any kinematic equations before?
 
yes i have
 
So, let's consider this problem in horizontal and vertical components. List the knowns and unknowns in each direction.
 
xcomp of 3 lb
a = 0
v= Vo
s=Vot + So = 0

Y comp of 3 lb
a= g
v= gt + Vo
s= gt^2/2 + Vot + So

xcomp of 5 lb
a = 0
v= Vo
s=Vot + So = 0= Vot + 1659

Y comp of 5 lb
a= g
v= gt + Vo
s= gt^2/2 + Vot + So = 0 =gt^2/2 + Vot + 875
 
Looks good to me :biggrin:, there's no need to consider the path of the 3lb particle since we are not asked to calculate it in the question. One small correction for your 5lb y component;

s= -gt^2/2 + Vot + So = 0 =-gt^2/2 + Vot + 875

So now you have two equations with two unknowns, I'm sure you know how to proceed from here.
 
Actually as said this is a momentum problem; so need to do all that ugly stuff =p. You're already past the "grunge" way of doing the problems.

It's like the idea of a ball striking another ball and both going off at different angles. This is a two dimensional collision essentially.

One piece will go off at angle gamma, the other will go off at angle beta.
 
I disagree, the problem is solvable if we consider it as a purely projectile motion problem, the fragment will behave as if it was a projectile given some initial horizontal velocity at some height above the ground [ignoring air resistance of course].
 
You can do it that way; however, it is a grunt method. The key thing was that it is momentum, so he/she should be learning it through momentum.

Such as you could go about projectile motion in how far something will land. Set it up into the x and y vectors, say that there's no acceleration in the x and do a lot. However, you can accomplish it through energy.

Momentum is the key in this and how to apply it is like with 2 dimensional collision. Same idea applies.
 
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