# Physics 30 question about conservation of momentum

• RRmy0440
In summary, the person throws a phaser at a velocity of 50.2km/h away from the spacecraft in order to get back. If the mass of the phaser is 5.3 kg, then the person has 0.28 minutes of extra air when they get back to the spacecraft.
RRmy0440

## Homework Statement

A space person is motionless a distance of 500m away from the safety of the spacecraft . The person has exactly 11.32min of air left and the person's mass is 103.2kg, including equipment. The person throws a phaser at a velocity of 50.2km/h away from the spacecraft in order to get back. I f the mass of the phaser is 5.3 kg, how many minutes of extra air does the person have when they get back to the spacecraft ？(Ans: 0.28min)

I attempted to solve it , but get the answer of 0.25min.

T

## Homework Equations

Conservation and momentum.
P=mv
Ft=mv
Impulse momentum theory

## The Attempt at a Solution

① m1v1+m2v2=m1v1'+m2v2'
v1'=42.97m/min

②t=500/(836.9m/min+42.97m/min) ＝0.57min

③d=v1't=42.97m/min *0.57min=24.42m
④d(remaining)=500m-24m=475.6m
⑤t=d/v=475.6m/ (42.97m/min)=11.07min
⑥t(remaining)=11.32-11.07=0.25min

I try to solve the question, but got the wrong answer. Could anyone help me out？ Thank you so much.

RRmy0440 said:
m1v1+m2v2=m1v1'+m2v2'
What values are you using for m1 and m2 here?
RRmy0440 said:
②t=500/(836.9m/min+42.97m/min) ＝0.57min

I would leave it in km/h till the latest possible moment (unless you've been taught differently).

haruspex said:
What values are you using for m1 and m2 here?

d=500m
t1=11.32min
m1=103.2kg
m2=5.3kg
v2'=50.2km/h=836.7m/min

verty said:
I would leave it in km/h till the latest possible moment (unless you've been taught differently).
If you would like to solve it with the unit km/h , please feel free to do that.

So do you think I am on the right track of solving this？

It's up to you what way you want to solve it. By the way, I got a different answer: 0.31. I don't think 0.28 is correct.

PS. Your way of doing it is confusing to me so I won't comment anymore.

RRmy0440 said:
m1=103.2kg
That is the mass of the astronaut including all equipment.
I still do not see any explanation for your equations.

Also, why do you add the two speeds?

Last edited:
verty said:
I got a different answer: 0.31. I don't think 0.28 is correct.
I get 0.28. Check you did not get -0.31. If you did, see the bold text in my post #8.

haruspex said:
I get 0.28. Check you did not get -0.31. If you did, see the bold text in my post #8.

haruspex said:
That is the mass of the astronaut including all equipment.
I still do not see any explanation for your equations.

Also, why do you add the two speeds?

Does it mean that the phaser is also included to have the total mass of 103.2kg？

For the reason why I added the two speeds together...I think I had been misled by another question I just done

Thank you guys, i get the right answer.

## 1. What is the law of conservation of momentum?

The law of conservation of momentum states that the total momentum of a closed system remains constant. This means that the total momentum before an event is equal to the total momentum after the event, as long as there are no external forces acting on the system.

## 2. How is momentum calculated?

Momentum is calculated by multiplying an object's mass by its velocity. The formula for momentum is p = m * v, where p is momentum, m is mass, and v is velocity. Momentum is measured in units of kilogram-meters per second (kg*m/s).

## 3. What is an elastic collision?

An elastic collision is a type of collision where both kinetic energy and momentum are conserved. In an elastic collision, the objects involved bounce off each other without any energy being lost or dissipated. This type of collision is often seen in billiard balls or other objects with high elasticity.

## 4. Can momentum be transferred between objects?

Yes, momentum can be transferred between objects through collisions or interactions. When two objects collide, some or all of one object's momentum can be transferred to the other object. This is the basis for the law of conservation of momentum.

## 5. How does conservation of momentum apply to real-world situations?

Conservation of momentum is a fundamental principle of physics and applies to many real-world situations. For example, it explains why a small car can be pushed by a larger car with equal force, and why a rocket moves forward when its exhaust is expelled backward. It is also important in understanding the dynamics of sports, such as when a pitcher throws a baseball or a hockey player shoots a puck.

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