A projectile exploding at the top of its trajectory

In summary, an object launched from the origin explodes into two fragments, one of mass m and the other of mass 3m. The smaller fragment lands back at the trajectory, while the larger fragment lands some distance away. Using the conservation of momentum and kinematic equations, it can be determined that the larger fragment will land at a distance of 3/4 x, where x is the distance between the two fragments before the explosion. The smaller fragment is blown back at the same speed it was traveling before the explosion, and the larger fragment is given a change in velocity equal to that of the smaller fragment. This allows for the determination of the distance the larger fragment will travel before landing.
  • #1
anban
20
0

Homework Statement



An object is launched from the origin. At the peak of it's trajectory, it explodes into two fragments, one of mass m and the other of mass 3m. The smaller fragment of mass m lands back at the trajectory.

How far from the origin does the second fragment land?

Homework Equations



Conservation of Momentum: P = mv
Conservation of Energy: KE = .5mv^2 only because explosion happens at singular height, so GPE irrelevant
Center of Mass: 1/M Ʃ mx
Kinematic equations

The Attempt at a Solution



The small mass m is at the origin, and the large mass 3m is some distance away. I will call this distance x.

Using the Center of Mass formula, (m(0)+3m(x))/(3m+m) = 3/4 x. So we know that the center of mass is 3/4 x. We also know that if the projectile did not explode, it would land at this spot.

Now, there are no times or velocities given. Using the kinematic equations and considering the even of the original projectile launching from the ground to the top of it's trajectory, we can say that:

x0=0
xf= 3/8 x
ax=0
ay=g
y0=0

This info combined with different kinematic equations gives these expressions:

0 = voy^2 +2gh
voy = gt
h = gt^2 + .5gt^2 = 3/2gt^2
0=g^2 t^ + 2gh
-g^2 t^4 / 2g = (3gt^2)/2
t^2 = 3
t = sqrt (3) This is the time it takes for our projectile to reach the top of it's path.

Now, I am ultimately out to get the x distance where the large mass is laying.
I think that I should continue manipulating the kinematic equations, but I am not sure that I have enough information.

I could fiddle around and get velocities for each of the pieces, I think. I could plug them into the conservation of momentum equation: but this wouldn't do me any good with finding a distance! Likewise with the conservation of energy.

Another useful piece of info: if m1 = 3m2, then according to cons. of momentum, I'm pretty sure that 3v1= v2.

Basically, I am not sure which direction to go in... I've been up and down this problem for a long time already, and I'm hoping that I am just forgetting some obvious point! Thanks in advance!
 
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  • #2
anban said:

Homework Statement



An object is launched from the origin. At the peak of it's trajectory, it explodes into two fragments, one of mass m and the other of mass 3m. The smaller fragment of mass m lands back at the trajectory.

How far from the origin does the second fragment land?

Homework Equations



Conservation of Momentum: P = mv
Conservation of Energy: KE = .5mv^2 only because explosion happens at singular height, so GPE irrelevant
Center of Mass: 1/M Ʃ mx
Kinematic equations

The Attempt at a Solution



The small mass m is at the origin, and the large mass 3m is some distance away. I will call this distance x.

Using the Center of Mass formula, (m(0)+3m(x))/(3m+m) = 3/4 x. So we know that the center of mass is 3/4 x. We also know that if the projectile did not explode, it would land at this spot.

Now, there are no times or velocities given. Using the kinematic equations and considering the even of the original projectile launching from the ground to the top of it's trajectory, we can say that:

x0=0
xf= 3/8 x
ax=0
ay=g
y0=0

This info combined with different kinematic equations gives these expressions:

0 = voy^2 +2gh
voy = gt
h = gt^2 + .5gt^2 = 3/2gt^2
0=g^2 t^ + 2gh
-g^2 t^4 / 2g = (3gt^2)/2
t^2 = 3
t = sqrt (3) This is the time it takes for our projectile to reach the top of it's path.

Now, I am ultimately out to get the x distance where the large mass is laying.
I think that I should continue manipulating the kinematic equations, but I am not sure that I have enough information.

I could fiddle around and get velocities for each of the pieces, I think. I could plug them into the conservation of momentum equation: but this wouldn't do me any good with finding a distance! Likewise with the conservation of energy.

Another useful piece of info: if m1 = 3m2, then according to cons. of momentum, I'm pretty sure that 3v1= v2.

Basically, I am not sure which direction to go in... I've been up and down this problem for a long time already, and I'm hoping that I am just forgetting some obvious point! Thanks in advance!

Like all good projectiles, the shell [before exploding] will have had a constant horizontal component to its velocity.
Also, the projectile got to maximum height in time T, and the bits will take an additional time T to get back to the ground.
In order to land at the origin, that means the small fragment has been blown back at exactly the same speed as it was traveling before the explosion - giving you the change in velocity of that fragment.
Equal impulses on the two fragments will show you by how much the velocity of the 3m mass changes, and thus how far - relative to the distance traveled so far - the 3m mass will go.
 

1. What is a projectile exploding at the top of its trajectory?

A projectile exploding at the top of its trajectory refers to a situation where a projectile, such as a bullet or a missile, explodes at the highest point of its flight path before falling back to the ground.

2. What causes a projectile to explode at the top of its trajectory?

There are a few possible causes for a projectile to explode at the top of its trajectory. It could be due to a malfunction or defect in the projectile itself, or it could be intentionally detonated by a remote trigger or a self-destruct mechanism.

3. Can a projectile explode at the top of its trajectory naturally?

No, a projectile cannot naturally explode at the top of its trajectory. Explosions require some sort of external force or energy to ignite, and a projectile's flight path does not generate enough energy to cause an explosion at its highest point.

4. Are there any real-life examples of projectiles exploding at the top of their trajectory?

Yes, there have been instances where projectiles have exploded at the top of their trajectory. One example is the self-destruct mechanism on military missiles, which are designed to detonate at a specific altitude to prevent them from falling into enemy hands.

5. What are the potential dangers of a projectile exploding at the top of its trajectory?

The dangers of a projectile exploding at the top of its trajectory depend on the type of projectile and the surroundings. If it is a military missile, it could cause damage or harm to nearby structures or people. In other cases, it could also result in a fire or release of hazardous materials.

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