# Help: Physics Homework Problem-Conservation of Linear Momentum

• shawonna23
In summary, by using the equation m1*v1 + m2*v2 = (m1+m2)vf and the fact that both skaters gain equal amounts of momentum from the compressed spring, we can determine that Al's mass is 109 kg. This is found by substituting the equation m1 + m2 = 168 kg and solving for m1 in terms of the given velocities. With the given velocity values, we can solve for the required mass of Al to be 109 kg.
shawonna23
Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.81 m/s, while Jo moves off in the opposite direction at a speed of 1.5 m/s. Assuming that friction is negligible, find Al's mass.

I am stuck on this problem. I think I have to use this equation: m1*v1 + m2*v2=(m1+m2)vf. But, I don't know how I would use it to solve this problem. Can someone please help me figure out how to solve this problem?

this is not a momentum conservation question becasue there is no collision of any sorts going on this question. There is however, energy conservation

the initial spring's energy is converted to the kinetic energies of the two skaters.

also think forces
on one side the spring is being compressed by a force of (168-M)a and it is turn exerts a force of k(x/2) in the oppsite direction.
From teh other side a force of Ma and spring force of k(x/2)
now you have one of their weight (pick one doesn't matter) to be 168 - M and other to be M, sub ito the energy conservation equation, and solve using the force relation

Last edited:
shawonna23 said:
Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.81 m/s, while Jo moves off in the opposite direction at a speed of 1.5 m/s. Assuming that friction is negligible, find Al's mass.

I am stuck on this problem. I think I have to use this equation: m1*v1 + m2*v2=(m1+m2)vf. But, I don't know how I would use it to solve this problem. Can someone please help me figure out how to solve this problem?
SOLUTION HINTS:
When released, compressed spring applies the same force F(t), which is a function of time, to both skaters over the same time interval, so that {∫ F(t) dt} is the same for both skaters. Therefore, both skaters gain equal amounts of momentum. Hence, their final momentums are related by:
m1*v1 = m2*v2
It is also known that:
m1 + m2 = (168 kg) ::: ⇒ m1 = 168 - m2
Place 2nd equation into the 1st, and solve for required "m" in terms of velocities "v" (evaluate using velocity values given in problem statement).

~~

Last edited:
Thanks for the help guys. I got the answer which was 109kg. It was correct! Thanks again!

## What is the concept of conservation of linear momentum?

The concept of conservation of linear momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time. 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.

## How is linear momentum defined?

Linear momentum is defined as the product of an object's mass and velocity. It is a vector quantity, meaning it has both magnitude and direction. The units of linear momentum are kg*m/s.

## What are the equations for calculating linear momentum?

The equation for calculating linear momentum is p = m * v, where p is momentum, m is mass, and v is velocity. In some cases, the equation may be written as p = mv, where the dot represents the mathematical operation of multiplication. Additionally, if there are multiple objects involved in a system, the total momentum can be calculated by adding the individual momenta together: p_total = m_1v_1 + m_2v_2 + ... + m_nv_n.

## How does the conservation of linear momentum apply to collisions?

The conservation of linear momentum applies to collisions by stating that the total momentum of the system before the collision is equal to the total momentum after the collision. This means that if two objects collide, the sum of their momenta before the collision is equal to the sum of their momenta after the collision.

## What are some real-life examples of the conservation of linear momentum?

Some real-life examples of the conservation of linear momentum include billiard balls colliding on a pool table, a tennis ball bouncing off a wall, or a rocket launching into space. In each of these examples, the total momentum of the system is conserved, even though individual momenta may change due to the interaction between objects or forces.

• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
720
• Introductory Physics Homework Help
Replies
35
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
23
Views
671
• Introductory Physics Homework Help
Replies
335
Views
9K
• Introductory Physics Homework Help
Replies
3
Views
929