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Devrin
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Center of Mass, Momentum, and energy problem, shell seperation
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)
Distance Formula
xf=xi+vt+at^2
Conservation of Momentum
mv1=m1u1+m2u2
Conservation of Energy
.5mv^2=.5m(u1)^2+.5m(u2)^2
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.
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.
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