# Center of Mass, Momentum, and you!

Center of Mass, Momentum, and energy problem, shell seperation

## Homework Statement

A projectile of mass 20.1 kg is fired at an angle of 65.0 degrees above the horizontal and with a speed of 79.0 m/s. At the highest point of its trajectory the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance.

A) How far from the point of firing does the other fragment strike if the terrain is level? (gravity is 9.8m/s^2)

## Homework Equations

Distance Formula
xf=xi+vt+at^2

Conservation of Momentum
mv1=m1u1+m2u2

Conservation of Energy
.5mv^2=.5m(u1)^2+.5m(u2)^2

## The Attempt at a Solution

ok, I started out by finding the total time it takes for the mass system to make it's parabola and land, which is 7.306s then I find the total x the mass system would have gone, which was 244m=79cos(65)*(7.306)
once I did this I started using momentum to find the initial speed of the second particle to get 66.77m/s. this is where I get stuck. I plug in my new data for the 2nd particle in the position formula, dividing time by two because it's already at it's peak, and I get 553m from it's original launch point. this is mastering physics and I don't even get a "you are close" error message. I am just completely stuck.

Last edited:

hage567
Homework Helper
I'm confused about the part where you say you are "dividing time by two because it's already at it's peak". That time is irrelevant to the fragment after the explosion. You need to find the new time it takes that fragment to reach the ground based on its new velocity after the explosion (Hint: you can get this time based on the first fragment that was initially at zero speed after the explosion).

To get the total distance, you must add the horizontal distance travelled by the large mass before the explosion, and what the second fragment travels after the explosion.

That's my take on it anyway.

Ok, you know the angle its fired at and the velocity. So, you take the component of velocity in the vertical direction (vsinx, where x is the angle), and using that you find the max height ($$u^2sin^2x=2gh$$).

Now, at this point you apply conservation of linear momentum along the horizontal, as the net force is zero along the horizontal. This gives you mvcosx=m(0)+m(v2).

From here you get your velocity v2, of the block that doesnt stop in mid air. You have the height, and the velocity. The distance the block travels in the horizontal direction in the time it takes for it to fall on the ground is what youre asked.