(Momentum) Two people on a platform

[Mr. Box]

Homework Statement

Two people are standing on a 2.0-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of air. One person throws a 6.0-kg ball to the other, who catches it. The ball travels horizontally. Excluding the ball, the total mass of the platform and people is 118 kg. Because of the throw, this 118-kg mass recoils. How far does it move before coming to rest again?

Homework Equations

Conservation of linear momentum:
(M1 * Vf1) + (M2 * Vf2) = (M1 *Vo1) + (M2 * Vo2)

The Attempt at a Solution

When the ball is thrown forwards, the platform should accelerate backwards because the air resistance is negligible. When the other person catches the ball the platform should move back to its original spot because the momentum of the ball is transferred to the platform, so I thought the answer was a displacement of zero, but its not. Can someone show me what wrong with my solution?

 

[NihalSh]

what you actually have to take into consideration is Center of Mass. If initially before throwing the ball, the system was at rest, then all throughout the system should be at rest in the x-direction regardless of whether the ball is moving or is caught by the other person because no external force is acting on the system in x-direction.

System includes ball+both people+platform. essentially everything on the platform

 

[Mr. Box]

I didn't even need to use linear momentum, thanks.

 

[NihalSh]

I didn't even need to use linear momentum, thanks.

yeah, true. glad to help

 

[Nickie]

Your understanding of the concept of momentum is correct, but there are a few factors that you may have overlooked in your solution.

Firstly, the platform is not completely stationary to begin with. It is floating on a cushion of air, which means it has some initial velocity. This initial velocity must be taken into account when calculating the final velocity of the platform after the ball is caught.

Secondly, the platform and people have a total mass of 118 kg, but only one person is throwing the ball. This means that the person throwing the ball will experience a recoil with a smaller mass, and the person catching the ball will experience a recoil with a larger mass. This will affect the final velocities of both individuals, which in turn will affect the final velocity of the platform.

To solve this problem, you can set up two equations using the conservation of momentum formula:

For the person throwing the ball:
(M1 * Vf1) + (M2 * Vf2) = (M1 * Vo1) + (M2 * Vo2)
where M1 is the mass of the person throwing the ball (let's call it m1), Vf1 is their final velocity, M2 is the mass of the platform and the other person (118 kg), Vf2 is the final velocity of the platform and other person, and Vo1 and Vo2 are their initial velocities (which we will assume to be zero).

For the person catching the ball:
(M1 * Vf1) + (M2 * Vf2) = (M1 * Vo1) + (M2 * Vo2)
where M1 is the mass of the person catching the ball (let's call it m2), Vf1 is their final velocity, M2 is the mass of the platform and the other person (118 kg), Vf2 is the final velocity of the platform and other person, and Vo1 and Vo2 are their initial velocities (which we will assume to be zero).

You now have two equations with two unknowns (Vf1 and Vf2). Solve for these variables, and you will be able to determine the final velocity of the platform and other person. From there, you can use the formula for displacement (d = Vf * t) to calculate how far the platform and people move before coming to rest again. Remember to take into account the initial velocity of the platform when calculating the time (t) it takes