Kinetic energy along with conservation of momentum

In summary: It just clutters up the forum. I'm going to close this thread, please continue in the original thread.
  • #1
Lolagoeslala
217
0

Homework Statement


a hockey player of mass 80 kg was skating at a velocity of 7.5 m/s [E 20 S] when he accidently hit a linesman who was just standing on the ice. The mass of the linesmass is 90 kg and his velocity after the collision was 3 m/s [E 30 N] find.
a) the velocity of the hockey player after the collision and
b) the kinetic energy lost during the collision as a percentage.

The Attempt at a Solution


given:
m1= 80 kg
m2 = 90 kg
v1 = 7.5 m/s [ E 20 S]
v1` = ?
v2= 0m/s
v2` = 3m/s [ E 30 N]

X component
m1v1 = m1v1` + m2m2`
(80kg)(7.047694656m/s[E]) = (80kg)(v1`) + (90kg)(2.598076211[E])
563.8155725 kgm/s [E] - 233.826859 kgm/s[E] = (80kg)(v1`)
329.9887135 [E] / 80kg = v1`
4.124858919 m/s [E] = v1`

y components
m1v1 = m1v1` + m2v2`
(80kg)(2.565151075m/s) = (80kg)(v1`) + (90kg)(1.5m/s[N])
205.212086kgm/s-135m/s[N] = (80kg)(v1`)
70.212086kgm/s/80kg = v1`
4.252651075m/s = v1`

4.124858919 m/s [E] + 4.252651075m/s = v1`
5.924 m/s [S 44° E] = v1`

b) Ektot = 1/2m1v1^2
Ektot= 1/2(80kg)(7.5m/s)^2
Ektot=2250 J

Ektot` = 1/2m1v1`^2 + 1/2m2v2`
Ektot` = 1/2(80kg)(5.924m/s)^2 + 1/2(90kg)(3m/s)^2
Ektot` =1403.75104 J + 405 kg J
Ektot` = 1808.75104 J

Ektot - Wdef = Ektot`
2250 J - Wdef = 1808.75104 J
Wdef = 441.24896 J

The Wdef is the energy lost during the collision

(441.248896 J / 2250 J) x 100%
19.6 %
 
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  • #2
Lolagoeslala said:

Homework Statement


a hockey player of mass 80 kg was skating at a velocity of 7.5 m/s [E 20 S] when he accidently hit a linesman who was just standing on the ice. The mass of the linesmass is 90 kg and his velocity after the collision was 3 m/s [E 30 N] find.
a) the velocity of the hockey player after the collision and
b) the kinetic energy lost during the collision as a percentage.

The Attempt at a Solution


given:
m1= 80 kg
m2 = 90 kg
v1 = 7.5 m/s [ E 20 S]
v1` = ?
v2= 0m/s
v2` = 3m/s [ E 30 N]

X component
m1v1 = m1v1` + m2m2`
(80kg)(7.047694656m/s[E]) = (80kg)(v1`) + (90kg)(2.598076211[E])
563.8155725 kgm/s [E] - 233.826859 kgm/s[E] = (80kg)(v1`)
329.9887135 [E] / 80kg = v1`
4.124858919 m/s [E] = v1`

y components
m1v1 = m1v1` + m2v2`
(80kg)(2.565151075m/s) = (80kg)(v1`) + (90kg)(1.5m/s[N])
205.212086kgm/s-135m/s[N] = (80kg)(v1`)
70.212086kgm/s/80kg = v1`
4.252651075m/s = v1`

4.124858919 m/s [E] + 4.252651075m/s = v1`
5.924 m/s [S 44° E] = v1`

b) Ektot = 1/2m1v1^2
Ektot= 1/2(80kg)(7.5m/s)^2
Ektot=2250 J

Ektot` = 1/2m1v1`^2 + 1/2m2v2`
Ektot` = 1/2(80kg)(5.924m/s)^2 + 1/2(90kg)(3m/s)^2
Ektot` =1403.75104 J + 405 kg J
Ektot` = 1808.75104 J

Ektot - Wdef = Ektot`
2250 J - Wdef = 1808.75104 J
Wdef = 441.24896 J

The Wdef is the energy lost during the collision

(441.248896 J / 2250 J) x 100%
19.6 %

This looks like a repeat of your earlier thread.

https://www.physicsforums.com/showthread.php?t=651165&highlight=hockey+player

Perhaps just a new attempt at a solution on your part.
 
  • #3
SammyS said:
This looks like a repeat of your earlier thread.

https://www.physicsforums.com/showthread.php?t=651165&highlight=hockey+player

Perhaps just a new attempt at a solution on your part.

yes but this time i tried to use the different orentrations since the previous one did not help me and the kinetic energy seemed to increase
 
  • #4
Please don't create multiple threads on the same topic.
 
  • #5




Based on the given information, it can be determined that the hockey player's velocity after the collision is 5.924 m/s [S 44° E] and the kinetic energy lost during the collision is 441.24896 J, which is equivalent to 19.6% of the total initial kinetic energy. This can be explained by the conservation of momentum and the transfer of energy during the collision. The mass and velocity of both the hockey player and the linesman were taken into account in order to calculate the final velocity and the energy lost. This demonstrates the importance of understanding and applying the principles of kinetic energy and conservation of momentum in real-life situations.
 

1. What is kinetic energy?

Kinetic energy is the energy possessed by an object due to its motion. It is a scalar quantity that depends on the mass and velocity of the object.

2. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics which states that in a closed system, the total momentum remains constant. This means that the total momentum before an event must be equal to the total momentum after the event, as long as there are no external forces acting on the system.

3. How is kinetic energy related to conservation of momentum?

Kinetic energy and conservation of momentum are closely related as both are dependent on the velocity of an object. In a system where there are no external forces, the total kinetic energy and total momentum will remain constant. This means that if the velocity of an object increases, both its kinetic energy and momentum will increase as well.

4. Can kinetic energy and momentum be lost in a system?

In a closed system, neither kinetic energy nor momentum can be lost. This is due to the conservation of momentum and energy laws. While energy and momentum can be transferred or transformed within the system, the total amount will always remain constant.

5. How is kinetic energy and momentum used in real-world applications?

Kinetic energy and momentum are utilized in various real-world applications, such as in sports, transportation, and engineering. In sports, athletes use their momentum and kinetic energy to perform actions like throwing or jumping. In transportation, the conservation of momentum is crucial in designing efficient and safe vehicles. In engineering, understanding kinetic energy and momentum is essential in designing machines and structures that can withstand forces and movement.

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