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.

 

[asteriatic]

As a scientist, it is important to carefully analyze all the given information and use the appropriate equations to solve the problem. In this case, you have correctly used the conservation of momentum and conservation of kinetic energy equations to determine the initial velocities of the blocks after the collision. However, to find the final velocities, we need to use the equations for conservation of momentum and conservation of kinetic energy after the collision.

Using the conservation of momentum equation, we can write:

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

And using the conservation of kinetic energy equation, we can write:

KEi = KEf
1/2(1.5)(10)^2 + 0 = 1/2(1.5)(V1)^2 + 1/2(6)(V2)^2
75 = 0.75(V1)^2 + 3(V2)^2

Now, we have two equations and two unknowns (V1 and V2). We can solve these equations simultaneously to find the values of V1 and V2. One way to do this is by substituting the value of V1 from the first equation into the second equation:

75 = 0.75(15-6V2) + 3(V2)^2
75 = 11.25 - 4.5V2 + 3(V2)^2
3(V2)^2 - 4.5V2 - 63.75 = 0

Solving this quadratic equation, we get two possible values for V2: 5 and -4.25. However, we know that the velocity of an object cannot be negative, so we can discard the negative value.

Therefore, the final velocity of the block with mass 6kg is 5m/s and the final velocity of the block with mass 1.5kg is 15-6(5)/1.5 = 5m/s.

b) To determine the distance the block with mass 6kg will travel up the incline, we can use the equation for conservation of energy:

PEi + KEi = PEf + KEf
0 + 1/2(6)(5)^2 = mgh + 1/2(6)(5)^2