A collision, help me double check this ?

[Jbreezy]

Homework Statement

A mass, m1 = 4.5 kg is released form rest 2.5 m above the ground on a inclined plane (35 degrees). m1 slides down incline and makes a smooth transition to flat plane then strikes m2 = 8.2 kg, which is at rest, calculate the speed of both masses after elastic collision and find how high m1 will go back up the incline.

Homework Equations

The Attempt at a Solution

Part 1.
Figure out how fast m1 is going after traveling down incline.
So,
mgh = 1/2 mv^2
manipulate for v and you get v = sqrt(2gh)
sub in values and you get v = 7m/s
Part 2 the collision
We will use
Vi2 = (m1- m2/ m1+m2) (vi1) + 2m2/m1+m2(v2I)
The V2i is 0 so it is gone.
Vi2 = (m1- m2/ m1+m2) (vi1)
We substitute in the values and we get -2.03m/s
For mass 2 we use.

V2II = (2m1)/ (m1+m2) (vi1) + (m1-m2/ m1+m2 )(V2i)
V2i once again drops out
V2II = (2m1)/ (m1+m2) (vi1)
Sub in values and we get 4.96m/s


Lastly get the height for mass one back up incline.
We know the velocity so we can use
1/2mvII = mgh
solve for h and get h = VII/2g
And I got .104m

I hope it is right please double check me.
Thanks

 

[barryj]

I agree with your answers

 

[Jbreezy]

thx dude