Conservation of momentum and energy

In summary, Gayle runs at a speed of 4.38 m/s and dives on a sled, which is initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.33 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. Their speed at the bottom of the hill is 12.6 m/s if the total vertical drop is 12.6 m. Gayle's mass is 46.5 kg, the sled has a mass of 6.00 kg and her brother has a mass of 32.4 kg.
  • #1
Zubz
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Homework Statement


Hi, new to the forums and I was wondering if someone could help me figure out what I am doing wrong. Thanx in advance.
Gayle runs at a speed of 4.38 m/s and dives on a sled, which is initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.33 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 12.6 m? Gayle's mass is 46.5 kg, the sled has a mass of 6.00 kg and her brother has a mass of 32.4 kg.

Homework Equations



K.E=1/2mv^2
P.E.=mgy
p=mv
Mgyle= 46.5 kg
M(gyle+sled)= 52.5 kg
M(gyle+sled+bro) = 84.9 KG

The Attempt at a Solution



Ok so what i did was i found out the initial speed of the sled using conservation of momentum.
p=p!(prime)
Mgyle*Ugyle = M(gyle+sled)V
V=3.89 m/s
then i used conservation of energy to find out the velocity of the sled at 5.35m.
Ebefore= Eafter
mgYo + 1/2mV1^2 = mgY1 + 1/2mV2^2
( 2(52.5)(9.81)(12.6) + (52.5*3.89^2) - (2*52.5*9.81*5.35) ) / 52.5 = v2^2
v2= 12.55 m/s
Now when i calculated this velocity. Should i include the brothers mass or do i find it after using conservation of momentum using the new velocity (12.55 m/s).
AND if i use conservation of momentum i get the new velocity to be 7.76 m/s (when brothers jumps on) at 5.35m.
After that, I use conservation of energy again to find final velocity.
Ebefore = Eafter
2MgY1 + MV2^2 = MV3^2
( (2*84.9*9.81*5.35) + (84.9*7.76^2) ) / 84.9 = v3^2 (USED NEW VELOCITY AFTER COLLISION WITH BROTHER)
v3 = 12.85 m/s but this isn't right. Can any1 tell me where i went wrong?
 
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  • #2
then i used conservation of energy to find out the velocity of the sled at 5.35m.
Ebefore= Eafter
mgYo + 1/2mV1^2 = mgY1 + 1/2mV2^2
( 2(52.5)(9.81)(12.6) + (52.5*3.89^2) - (2*52.5*9.81*5.35) ) / 52.5 = v2^2

After she has descended a vertical distance of 5.33 m,

I believe I have found your problem. When the sled descends a vertical distance of 5.33 m, the sled is a vertical distance of (12.6 - 5.33) m above the ground. You should come up with about 10.94 m/sec for the speed at that point before the brother hops on.

Now when i calculated this velocity. Should i include the brothers mass or do i find it after using conservation of momentum using the new velocity (12.55 m/s).
Find the velocity of the sled without the brother's mass at (12.6 - 5.33) m above the ground. Then use conservation of momentum with the new velocity to get the velocity after the brother gets on. Then use the result to calculate the total energy. The reasoning being that the sled is already traveling at a certain velocity before the brother gets on. When the brother jumps on, he not only changes the configuration of the potential and kinetic energy, but he adds to the total energy too. In order to figure out the new total energy, you need to find both the new potential energy that he adds and the new kinetic energy of the sled after he jumps on, but in order to calculate the new kinetic energy, you need to use the conservation of momentum.
 
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  • #3
thank you, worked perfectly.
 

FAQ: Conservation of momentum and energy

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, regardless of any external forces acting on the system. This means that in any interaction between objects, the total momentum before the interaction is equal to the total momentum after the interaction.

2. How does the conservation of momentum apply to everyday life?

The conservation of momentum can be seen in everyday life through various examples such as a car collision, a ball bouncing, or even walking. In a car collision, the momentum of the two cars before the crash is equal to the momentum of the two cars after the crash. In the case of walking, when we take a step, our foot exerts a force on the ground, and the ground exerts an equal and opposite force on our foot, resulting in a change in momentum.

3. What is the difference between conservation of momentum and conservation of energy?

The conservation of momentum and conservation of energy are two separate laws, but they are closely related. The conservation of momentum applies to the motion of an object or system, while the conservation of energy applies to the total energy of the system. In other words, momentum is a measure of how an object is moving, while energy is a measure of the object's ability to do work.

4. Can the conservation of momentum and energy be violated?

No, the conservation of momentum and energy are fundamental laws of physics and cannot be violated. These laws have been repeatedly tested and observed to hold true in all physical interactions. However, in some cases, it may seem like momentum or energy are not conserved, but this is due to external forces or factors that are not accounted for in the system.

5. How does the conservation of momentum and energy impact the environment?

The conservation of momentum and energy play a crucial role in maintaining the balance of energy in the environment. For example, the energy from the sun is conserved and transferred through various processes such as photosynthesis, which is essential for the growth of plants and maintaining the balance of ecosystems. Additionally, the conservation of momentum is also important in understanding and predicting the motion of objects such as planets and stars in the universe.

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