Elastic collision/ projectile motion

[rugbygirl2]

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).
I have
m1= 4kg, and a initial vx of 2 m/s
m2= 8kg, and a initial vx of 1 m/s
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!

 

[schroder]

This probably belongs in the homework section. You should show how you calculated your velocities. Hint: The values may be correct, but recheck which velocity is for which mass, and be alert for a rebound after collision. Try using simultaneous equations involving the conservation of momentum and the conservation of energy.

 

[astrokreng]

I would approach this problem by first analyzing the given information and identifying the variables that are known and unknown. From the given information, we know the masses of the objects (m1=4kg, m2=8kg), their initial velocities (m1=3m/s, m2=0m/s), and the length of the table (2m). The unknown variables are the height of the table, the time of flight, and the distance traveled by each object (delta x).

To solve for the distance traveled by each object, we can use the principles of conservation of momentum and conservation of energy. In an elastic collision, the total momentum and total kinetic energy of the system are conserved. This means that the sum of the initial momenta of the objects (m1v1 + m2v2) is equal to the sum of the final momenta (m1v1' + m2v2') and the sum of the initial kinetic energies (1/2m1v1^2 + 1/2m2v2^2) is equal to the sum of the final kinetic energies (1/2m1v1'^2 + 1/2m2v2'^2).

Using these principles, we can set up the following equations:

m1v1 + m2v2 = m1v1' + m2v2'

1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2

Substituting the known values, we get:

(4kg)(3m/s) + (8kg)(0m/s) = (4kg)(2m/s) + (8kg)(1m/s)

(1/2)(4kg)(3m/s)^2 + (1/2)(8kg)(0m/s)^2 = (1/2)(4kg)(2m/s)^2 + (1/2)(8kg)(1m/s)^2

Solving these equations, we get:

v1' = 2m/s
v2' = 1m/s

Now, to find the distance traveled by each object, we can use the equation for projectile motion:

delta x = v0t + 1/2gt^2

Where v0 is the initial