Determine the velocity of the two blocks right after the collision

[cupcaked]

Here's a (crude) drawing: http://i44.tinypic.com/2mg9ts2.png
A block with mass 1.5kg is pushed against a spring such that the spring compression is 0.2m. The block is then released from rest and is observed to have a velocity of 10m/s once it is released by the spring. The block then collides elastically with another block of mass 6 kg. The incline makes an angle 37 with the horizontal and is frictionless.

a) Determine the velocity of the two blocks right after the collision.
b) Determine the distance the block of mass 6kg will travel up the incline.
c) Depending on the direction that the block with mass 6kg moves after the collision, determine the maximum compression of the spring OR the distance it travels up the incline.


a) Help! I did all the work but I ended up with the initial velocities instead of getting the final velocities, how am I supposed to do this?

Pi = Pf
(1.5)(10) + 0 = (1.5)(V1) + (6)(V2)
15 = (1.5)(V1) + (6)(V2)
(1.5)(V1) = 15 - ^V2
V1 = -(6V2+15)/(1.5)

KEi = KEf
1/2(1.5)(10)^2 + 0 = 1/2(1.5)(V1)^2 + 1/2(6)(V2)^2
75 = 0.75(-6V2+15/1.5)^2 + 3V2^2
-3V2^2 + 75 = 0.75(-6V2+15/1.5)^2
-4V2^2 + 100 = (-6V2+15/1.5)^2
-4V2 + 10 = (-6V2+16/1.5)
-6V2 + 15 = -6V2 + 15

so V2 = 0

V1 = (-6(0)+15/1.5)
so V1 = 10

but those are the initial velocities... How do i find the final velocities?

 

[cepheid]

Here's a (crude) drawing: http://i44.tinypic.com/2mg9ts2.png
A block with mass 1.5kg is pushed against a spring such that the spring compression is 0.2m. The block is then released from rest and is observed to have a velocity of 10m/s once it is released by the spring. The block then collides elastically with another block of mass 6 kg. The incline makes an angle 37 with the horizontal and is frictionless.

a) Determine the velocity of the two blocks right after the collision.
b) Determine the distance the block of mass 6kg will travel up the incline.
c) Depending on the direction that the block with mass 6kg moves after the collision, determine the maximum compression of the spring OR the distance it travels up the incline.


a) Help! I did all the work but I ended up with the initial velocities instead of getting the final velocities, how am I supposed to do this?

Pi = Pf
(1.5)(10) + 0 = (1.5)(V1) + (6)(V2)
15 = (1.5)(V1) + (6)(V2)
(1.5)(V1) = 15 - ^V2
V1 = -(6V2+15)/(1.5)

KEi = KEf
1/2(1.5)(10)^2 + 0 = 1/2(1.5)(V1)^2 + 1/2(6)(V2)^2
75 = 0.75(-6V2+15/1.5)^2 + 3V2^2
-3V2^2 + 75 = 0.75(-6V2+15/1.5)^2
-4V2^2 + 100 = (-6V2+15/1.5)^2
-4V2 + 10 = (-6V2+16/1.5)
-6V2 + 15 = -6V2 + 15

so V2 = 0

V1 = (-6(0)+15/1.5)
so V1 = 10

but those are the initial velocities... How do i find the final velocities?

See the error in red above.\sqrt{-4v_2^2 + 100} \neq -4v_2 + 10

I would recommend solving for v2 by isolating it on one side of the equation.

 

[cupcaked]

I'm really messing up my math here or something...

KEi = KEf
1/2(1.5)(10)^2 + 0 = 1/2(1.5)(V1)^2 + 1/2(6)(V2)^2
75 = 0.75(-6V2+15/1.5)^2 + 3V2^2
75 = 0.75(-4V2+10)^2 + 9V2
75 = 0.75(16V2+100) + 9V2
75 = 12V2 + 75 + 9V2
75 = 21V2 + 75
0 = 21V2 ...? zero again... blehhh!

 

[cepheid]

I'm really messing up my math here or something...

KEi = KEf
1/2(1.5)(10)^2 + 0 = 1/2(1.5)(V1)^2 + 1/2(6)(V2)^2
75 = 0.75(-6V2+15/1.5)^2 + 3V2^2
75 = 0.75(-4V2+10)^2 + 9V2
75 = 0.75(16V2+100) + 9V2
75 = 12V2 + 75 + 9V2
75 = 21V2 + 75
0 = 21V2 ...? zero again... blehhh!

Where did this stuff in red come from? Why is it suddenly 4v₂ instead of 6v₂? Where did the 9 come from? Why is the v₂ that is multiplying the 9 no longer squared?

Note that (-6v_2 + 10)^2 = (-6v_2 + 10)(-6v_2 + 10)So you're going to have to expand that out.