A projectile exploding at the top of its trajectory

Click For Summary
SUMMARY

The discussion focuses on a physics problem involving a projectile that explodes at the peak of its trajectory into two fragments, one with mass m and the other with mass 3m. The smaller fragment lands back at the origin, while the larger fragment's landing distance from the origin is to be determined. Key principles applied include Conservation of Momentum, Conservation of Energy, and the Center of Mass formula. The time taken for the projectile to reach its peak is calculated as t = sqrt(3), which is crucial for further calculations regarding the landing distance of the larger fragment.

PREREQUISITES
  • Understanding of Conservation of Momentum in physics
  • Familiarity with Kinematic equations for projectile motion
  • Knowledge of Center of Mass calculations
  • Basic principles of Conservation of Energy
NEXT STEPS
  • Explore detailed applications of Conservation of Momentum in multi-body systems
  • Study Kinematic equations in-depth, focusing on projectile motion
  • Learn about the implications of Center of Mass in dynamic systems
  • Investigate the effects of explosions on motion and trajectory analysis
USEFUL FOR

Students studying physics, particularly those focused on mechanics and dynamics, as well as educators looking for illustrative examples of projectile motion and conservation laws in action.

anban
Messages
20
Reaction score
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!
 
Last edited:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 38 ·
2
Replies
38
Views
7K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
15
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
21
Views
4K
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K