Conservation of Momentum of Rocket Exploding after Takeoff

AI Thread Summary
The discussion focuses on the conservation of momentum in a rocket explosion scenario. Calculations revealed a pre-explosion speed of 20.0 m/s and a travel distance of 510 m for the lighter piece post-explosion. Participants noted that the actual masses of the rocket components were irrelevant for the calculations, as only their ratio mattered. There was uncertainty regarding the speed and direction of the heavier portion, with consensus that it would be downward. Overall, the conversation highlights key calculations and clarifications in understanding momentum conservation during the rocket's explosion.
hisiks
Messages
6
Reaction score
0
Homework Statement
A 1500 kg weather rocket accelerates upward at 10.0 m/s^2. It explodes 2.00 s after liftoff and
breaks into fragments. One piece is twice as massive as the other. Photos reveal that the
lighter fragment travelled straight up and reached a maximum height of 530 m. What was the
speed and direction of the heavier fragment just after the explosion?
Relevant Equations
pi = pf
m1v1i + m2v1i = m1v2f + m2v2f
vf = vi +at
d = vit + 1/2at^2
vf^2 = vi2 + 2ad
PEi + KEi = PEf + KEf
-Solved for vf using equation 3 to get 20.0m/s (speed before explosion) then solved for the distance to reach the explosion using equation 4, to get 20.0m, which felt wrong having the same numbers but that may just be coincidence.
-Found the distance travelled of the lighter piece using 530m - 20.0m to get 510m travelled after the explosion.
-Stuck trying to use equation 6 to find inital velocity.
Stuck with the remainder of the question.

EDIT: Found masses using Mtotal = m1 + m2, & m2 = 2m1, thefore m1 = 500kg and m2 = 1000kg.
 
Last edited:
Physics news on Phys.org
hisiks said:
that may just be coincidence
It's because the time is 2 seconds and ##\frac 122^2=2##.
hisiks said:
Found masses
The actual masses are irrelevant, unless another part of the question asked for them. The given ratio is all you need.
What did you get for the speed and direction of the heavier portion?
 
haruspex said:
It's because the time is 2 seconds and ##\frac 122^2=2##.

The actual masses are irrelevant, unless another part of the question asked for them. The given ratio is all you need.
What did you get for the speed and direction of the heavier portion?
This is what I came up with for a speed, however I'm not particularly confident in it. As for the direction, would it just be down?
 

Attachments

  • image.jpg
    image.jpg
    37.8 KB · Views: 147
hisiks said:
This is what I came up with for a speed, however I'm not particularly confident in it. As for the direction, would it just be down?
Looks right.
 
haruspex said:
Looks right.
Would you agree the direction would just be down as well?
 
hisiks said:
Would you agree the direction would just be down as well?
yes
 
Thread 'Minimum mass of a block'

Thread 'Minimum mass of a block'

Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
View full post »

Thread 'Electric field due to charges between 2 parallel infinite planes'

TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
View full post »
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
View full post »

Similar threads

Solving Elastic Collision of Two Balls: Theory & Solutions
Replies
15
Views
3K
Momentum in a perfectly inelastic collision
Replies
6
Views
2K
Conservation of momentum - with understanding
Replies
2
Views
2K
Confusion with "explosive" part, Conservation of Momentum
Replies
6
Views
2K
1D Elastic Collision between an Elephant and a Fly
Replies
6
Views
2K
An exploding cannon shell
Replies
4
Views
744
Elastic head-on collision problem
Replies
8
Views
2K
Conservation of Mechanical Energy - Which Equation To Use?
Replies
9
Views
1K
Elastic Collision and Momentum of Ice Skaters
Replies
16
Views
3K
Conservation of momentum & energy
Replies
11
Views
2K

Hot Threads

Recent Insights

Back
Top