Conservation of Momentum involving a space ship blowing up into three pieces

In summary, a spaceship of mass 2.30×10^6kg traveling at a speed of 5.40×10^6m/s breaks into three pieces when the antimatter reactor fails. One piece with a mass of 5.20×10^5kg is blown backward at a speed of 2.20×10^6m/s, while a second piece with a mass of 7.80×10^5kg continues forward at 1.20×10^6m/s. The mass of the third piece is calculated to be 1.00×10^6kg and its velocity is -1.69×10^-2m/s.
  • #1
SherBear
81
0

Homework Statement


A spaceship of mass 2.30×10^6kg is cruising at a speed of 5.40×10^6m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.20×10^5kg, is blown straight backward with a speed of 2.20×10^6m/s. A second piece, with mass 7.80×10^5kg, continues forward at 1.20×10^6m/s.

What is the speed of the third piece? in m/s


Homework Equations


M3=m3=M-m1+m2

Conservation of Momentum says:

m1v1+m2v2+m3v3=MV


The Attempt at a Solution



To get the weight of the third piece i used
M3=m3=M-m1+m2=
2.30*10^6kg - 5.20*10^5 kg + 7.80*10^5kg= 2.56*10^6 kg

Then
m1v1+m2v2+m3v3=MV

(5.20*10^5kg)(2.20*10^6)+(7.80*10^5kg)(1.20*10^6 m/s)+(2.56*10^6 kg) V3 = (2.30*10^6kg)(5.40*10^6ms)=

This may sound dumb but I don't know how to do this equation

I have 1.44*10^12 + 9.36*10^11 + 2.50*10^6

2.0800025*10^12 = 1.242*10^13

Then do I divide those 2, if so I get 1.67*10^-1 and it's wrong

What am I doing wrong?

Thank you
 
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  • #2
SherBear said:
To get the weight of the third piece i used
M3=m3=M-m1+m2=
2.30*10^6kg - 5.20*10^5 kg + 7.80*10^5kg= 2.56*10^6 kg
Shouldn't that be: m3 = M - m1 - m2 ?
 
  • #3
Doc Al said:
Shouldn't that be: m3 = M - m1 - m2 ?

I don't know, that piece is going to the right which is to the east and I'm calling that way positive?
 
  • #4
SherBear said:
I don't know, that piece is going to the right which is to the east and I'm calling that way positive?

It is positive just a mistake, I added it in the equation.
 
  • #5
SherBear said:
I don't know, that piece is going to the right which is to the east and I'm calling that way positive?
The direction of motion doesn't affect the mass. (It affects momentum, of course.)
 
  • #6
Doc Al said:
The direction of motion doesn't affect the mass. (It affects momentum, of course.)

I tried subtracting that and I have
1*10^6kg for mass 3

When i plugged it into the conservation I still get 1.67*10^-1 ?
 
  • #7
SherBear said:
I tried subtracting that and I have
1*10^6kg for mass 3
Good.

When i plugged it into the conservation I still get 1.67*10^-1 ?
How can you get the same answer with different numbers?

SherBear said:
Then
m1v1+m2v2+m3v3=MV

(5.20*10^5kg)(2.20*10^6)+(7.80*10^5kg)(1.20*10^6 m/s)+(2.56*10^6 kg) V3 = (2.30*10^6kg)(5.40*10^6ms)=
Two problems:
(1) As already pointed out, you have the wrong value for the mass of the third piece.
(2) You didn't incorporate the direction of motion. Things that go forward should have + velocity; things that go backward should have -.
 
  • #8
Doc Al said:
Good.


How can you get the same answer with different numbers?


Two problems:
(1) As already pointed out, you have the wrong value for the mass of the third piece.
(2) You didn't incorporate the direction of motion. Things that go forward should have + velocity; things that go backward should have -.

Ok m1 is positive and v1 is negative because it's going backwards?
m2 positive because it's weight, and m2 is positive because it's going forwards?
m3 doesn't matter because it's weight...then solve to get v3?
 
  • #9
SherBear said:
Ok m1 is positive and v1 is negative because it's going backwards?
m2 positive because it's weight, and m2 is positive because it's going forwards?
m3 doesn't matter because it's weight...then solve to get v3?
Almost: Masses are always positive, regardless of direction.

v1 is negative, v2 is positive. v3 you will solve for.
 
  • #10
Doc Al said:
Almost: Masses are always positive, regardless of direction.

v1 is negative, v2 is positive. v3 you will solve for.

Ok good, the only value i had to change was v1 to negative because it is going to the left or to the west.

it made V1 value negative

(5.20*10^5)(-1.144*10^12)+(9.36*10^11)+(1*10^6)=(2.30*10^6kg)(5.40*10^6m/s)

i get -1.69*10^-2, is this correct?
 
  • #11
SherBear said:
Ok good, the only value i had to change was v1 to negative because it is going to the left or to the west.

it made V1 value negative

(5.20*10^5)(-1.144*10^12)+(9.36*10^11)+(1*10^6)=(2.30*10^6kg)(5.40*10^6m/s)

i get -1.69*10^-2, is this correct?

oopse disregard the (5.20*10^5), my mistake
 
  • #12
SherBear said:
Ok good, the only value i had to change was v1 to negative because it is going to the left or to the west.

it made V1 value negative

(5.20*10^5)(-1.144*10^12)+(9.36*10^11)+(1*10^6)=(2.30*10^6kg)(5.40*10^6m/s)

i get -1.69*10^-2, is this correct?

I was able to use a more simple way and now it's correct, thanks!
 

1. How does the conservation of momentum apply to a space ship blowing up into three pieces?

The conservation of momentum states that the total momentum of a closed system remains constant. In the case of a space ship blowing up into three pieces, the total momentum of the system before and after the explosion must be equal.

2. What factors affect the conservation of momentum in this scenario?

The mass and velocity of each individual piece of the space ship will affect the conservation of momentum. The direction and angle at which the pieces are propelled after the explosion will also play a role.

3. Does the conservation of momentum also apply to the debris and surroundings of the explosion?

Yes, the conservation of momentum applies to the entire system, including any debris and surroundings that may be affected by the explosion. The total momentum of the system, including all objects and particles, will remain constant.

4. How can we calculate the momentum of each piece after the explosion?

The momentum of each piece can be calculated by multiplying its mass by its velocity. Using the law of conservation of momentum, we can also determine the velocity of each piece based on the known momentum of the original space ship before the explosion.

5. What real-life applications can we see of the conservation of momentum involving explosions in space?

The conservation of momentum is a fundamental principle in physics and is applied in various fields such as rocket propulsion, asteroid collisions, and satellite launches. Understanding and accurately predicting the momentum of objects in space can help us better navigate and explore our universe.

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