Conservation of momentum question

[frasifrasi]

Question:
A shell is fired with an initial velocity of 20 m/s at an angle of 60° above the horizontal.
At the top of its trajectory, the shell explodes into two fragments of equal mass. One
fragment, whose speed immediately after the explosion is zero, falls vertically. How far
from the gun does the other fragment land? Assume that the terrain is level and air
resistance is negligible.

Should I do this with conservation of momentum for each dimmension? I don't know how I would get a distance from this...
So basically I am having a hard time getting the problem set up, I appreciate if anyone can help me with this.

 

[Doc Al]

Questions to guide you:
- Where is the top of the shell's trajectory? (Use coordinates where the gun is at the origin.)
- What is the shell's velocity and momentum just before it explodes?
- What is the initial momentum and velocity of the two fragments?
- Treating the "other" fragment as a projectile, where does it land?

 

[ogb p]

Yes, you can use conservation of momentum for each dimension to solve this problem. Here's how you can set it up:

Let's label the fragments as A and B, with A being the fragment that falls vertically and B being the other fragment.

First, we need to find the initial momentum of the shell before it explodes. We know that the initial velocity is 20 m/s and the mass of the shell is equal to the mass of the two fragments combined. Therefore, the initial momentum (p) is:

p = m*v = (2m)*20 = 40m

Where m is the mass of each fragment.

Next, we can use conservation of momentum in the vertical direction to find the final momentum of fragment A. Since it falls vertically, its initial vertical momentum is zero. After the explosion, its final vertical momentum is also zero. Therefore, we can set up the following equation:

0 = m*vA - m*vA

Where vA is the vertical velocity of fragment A after the explosion.

Solving for vA, we get:

vA = 0 m/s

This means that fragment A will have no horizontal velocity and will fall straight down.

Now, we can use conservation of momentum in the horizontal direction to find the horizontal velocity of fragment B after the explosion. Since there is no external force acting on the system, the total momentum in the horizontal direction remains constant. Therefore, we can set up the following equation:

40m = m*vB

Where vB is the horizontal velocity of fragment B after the explosion.

Solving for vB, we get:

vB = 40 m/s

Now, we can use the equation for projectile motion to find the horizontal distance (d) traveled by fragment B:

d = vB*t

Where t is the time it takes for fragment B to land. We can find this time by using the equation for projectile motion in the vertical direction:

y = y0 + v0*t + (1/2)*a*t^2

Where y0 is the initial vertical position (which is zero), v0 is the initial vertical velocity (which is also zero), a is the acceleration due to gravity (9.8 m/s^2), and y is the vertical distance traveled (which is also zero).

Solving for t, we get:

t = √(2*y/a)

Since the vertical distance traveled is zero, we get:

t = 0 s