# Momentum of cannon ball homework

Aggie
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 jsut 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?

## Answers and Replies

Staff Emeritus
Gold Member
Convervation of energy is not required for this question. Have you met any kinematic equations before?

Aggie
yes i have

Staff Emeritus
Gold Member
So, lets consider this problem in horizontal and vertical components. List the knowns and unknowns in each direction.

Aggie
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

Staff Emeritus
Gold Member
Looks good to me , 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.

AngeloG
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.

Staff Emeritus
Gold Member
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].

AngeloG
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.