Elastic Collision: Solving for Delta x

[rugbygirl2]

Homework Statement

I have a 4kg mass (m1) moving at 3m/s towards a 8kg mass (m2) at rest on a 2m long frictionless table that is of unknown height. I solved for the velocities after the collsion m1= 2m/s and m2= 1m/, suppose the two masses were placed so that they left the edges of the table in opposite directions at the same time, ie the collision happened so that m2 was placed 0.66667m from the edge. What is the distance that each travels from the base of the table (delta x).

Homework Equations

m1= 4kg, and a initial vx of 2 m/s
m2= 8kg, and a initial vx of 1 m/s

The Attempt at a Solution

I do not know the height or time of flight so any way I substitute equations I get 2 unknowns. Any help or ideas would be appreciated!

 

[tiny-tim]

Welcome to PF!

Hi rugbygirl2! Welcome to PF!

You should have two equations, which should be enough for two unknowns.

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[thomate1]

To solve for delta x, we can use the conservation of momentum and conservation of energy equations for an elastic collision. The conservation of momentum equation states that the total momentum before the collision is equal to the total momentum after the collision. In this case, we have two objects colliding, so we can write this equation as:

m1v1i + m2v2i = m1v1f + m2v2f

where m1 and m2 are the masses of the two objects, v1i and v2i are their initial velocities before the collision, and v1f and v2f are their final velocities after the collision.

The conservation of energy equation states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. We can write this equation as:

1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2

Now, we can use these two equations to solve for delta x. First, we need to find the final velocities after the collision. We are given that m1v1f = 2m/s and m2v2f = 1m/s. We can use these values to solve for v1f and v2f:

v1f = 2m/s / m1 = 2m/s / 4kg = 0.5m/s
v2f = 1m/s / m2 = 1m/s / 8kg = 0.125m/s

Now, we can plug these values into the conservation of energy equation to solve for delta x:

1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2

1/2(4kg)(2m/s)^2 + 1/2(8kg)(1m/s)^2 = 1/2(4kg)(0.5m/s)^2 + 1/2(8kg)(0.125m/s)^2

16J + 4J = 1J + 0.5J

19J = 1.5J

Delta x = 19J /