# Center of Mass, Momentum, and you

• Devrin
In summary, the problem involves a projectile being fired at an angle with a certain velocity, which explodes into two fragments with equal mass at the highest point of its trajectory. Using conservation of momentum and energy, the initial speed of the second fragment is found and the distance it travels before striking the ground is calculated.
Devrin
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:
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 traveled 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 doesn't 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.

## What is the center of mass?

The center of mass is the point in an object or system where the mass can be considered to be concentrated. It is the average position of all the mass in the object or system.

## How do you calculate the center of mass?

The center of mass can be calculated by taking the weighted average of the positions of all the individual masses in the object or system. This can be done using the formula: xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where x represents the position and m represents the mass.

## What is momentum?

Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

## How is momentum conserved?

Momentum is conserved in a closed system, meaning that the total momentum of the system remains constant. This means that the sum of the momenta of all the objects in the system before a collision or interaction must be equal to the sum of the momenta after the collision or interaction.

## What is the relationship between momentum and velocity?

Momentum and velocity are directly proportional to each other. This means that an object with a higher velocity will have a higher momentum. However, the direction of the velocity and momentum may be different, as momentum is a vector quantity while velocity is a scalar quantity.

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