Uncertainty about answer for Conservation of Momentum Problem

In summary, a man with a total mass of 60kg skates on frictionless ice at an initial speed of +2 m/s and fires a 100 gram bullet with a speed of +500 m/s from a gun. The man's speed after firing the gun is +1.17 m/s. The initial total energy of the system is 120 J and the final total energy is 12541 J. This is an example of an inelastic collision as the KE before and after are not the same, with the extra KE coming from the chemical energy of the gunpowder.
  • #1
zozajiek
3
0

Homework Statement



A man holding a gun ((mass of man)+(mass of gun)+(mass of bullet)=60kg)) skates on frictionless ice at an initial speed of +2 m/s and fires a 100 gram bullet with a speed of +500 m/s from the gun.

a) What is the speed of the man after firing the gun? (be sure to include +/- direction)

b) What is the initial total energy of the man-gun-bullet system (before the bullet is fired)? What is the final total energy of the man-gun-bullet system (just after the bullet is fired)?

c)is this an example of an elastic, inelastic, or perfectly inelastic collision, and why? I don't understand this concept very well.


Homework Equations



Pi=Pf

M1V1i + M2V2i = M1V1f + M2V2f

Total KE (before)= .5MV^2

Total KE (after)= .5M1V1^2 + .5M2V2^2


The Attempt at a Solution



I think this is correct, but I'm not completely sure.
I'm not sure if its elastic or inelastic, I have trouble understanding that stuff.

60*2=59.9*v+.1*500
v=1.17 m/s

energy before is
.5*60*4 J
120 J

after
.5*.1*500^2+.5*59.9*1.17^2 J
12541 J
 
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  • #2
Recheck value of energy after.
 
  • #3
sorry it took me so long to get back on, what did I do wrong?
 
  • #4
You didn't. I did while calculating. Sorry.

This is an example of an inelastic collision, because the KE before and after are not the same. The extra KE is coming from the chemical energy of the gunpowder, which was converted to mechanical energy.
 
  • #5
=) thank you very much
 

1. What is conservation of momentum?

Conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system remains constant, unless acted upon by external forces. In other words, the total momentum before an event or interaction must equal the total momentum after the event or interaction.

2. Why is conservation of momentum important?

Conservation of momentum is important because it helps us understand and predict the behavior of objects in motion. It allows us to accurately calculate and analyze the effects of forces and interactions between objects. This law is also crucial in many real-world applications, such as in engineering and transportation systems.

3. What are some examples of conservation of momentum in everyday life?

Some examples of conservation of momentum in everyday life include a moving car colliding with another car and a rocket launching into space. In both cases, the total momentum before and after the event remains constant, demonstrating the principle of conservation of momentum.

4. How is conservation of momentum related to Newton's Third Law of Motion?

Conservation of momentum is closely related to Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. This means that when two objects interact, they exert equal and opposite forces on each other, resulting in a transfer of momentum. Therefore, the total momentum of the system remains constant.

5. What factors can affect conservation of momentum in a system?

The conservation of momentum in a system can be affected by external forces, such as friction or air resistance, which can cause a change in the total momentum of the system. Additionally, if the system is not closed, such as in an open system with external forces acting on it, the total momentum may not remain constant.

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