- #1

Temple1998

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## Homework Statement

:Two blocks with masses m1=3kg and m2=4.5kg are moving on a platform with velocities v1 and v2, respectively. The platform is inclined with θ=30ο and is frictionless. Mass m2 has a very stiff spring with constant k=3000kg/s2 attached as shown in the figure. The two blocks collide and the moment the spring compresses to its maximum compression xmax, the velocity of the blocks is common and has magnitude V (not known). You are given the following distances: AB=3cm, AC=8cm, AE=9cm, AF=12cm. The three pictures below correspond to the moment:

- right before touches the spring where the two objects move with v1 and v2 (This picture shows the blocks at positions A and B).
- the spring is at maximum compression and the system moves with common V (This picture shows the blocks at positions C and D).
- the two blocks lost contact and move with separate speeds (This picture shows the blocks at positions E and F).

b) the maximum compression max,

c) the velocity of each block after they lost contact.(Hint: you may construct the equations without numerical substitution)

[/B]

## Homework Equations

[/B]

m1v1=m2v2

E1 = E2

m1v1 + m2v2 = m1v1' + m2v2'

## The Attempt at a Solution

I cannot figure out how/whether the distances contribute to the problem since the blocks are moving up a slope. My professor allows us to answer the questions without substituting the numbers in.

**a)**

I used the Conservation of Momentum equation: m1v1 + m2v2 = (m1 + m2)V and rearranged the equation to solve for V. All values were known except V.

This gave me a final answer of

**V = (m1v1 + m2v2)/(m1 + m2).**

b)

b)

I used the equation: (1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)V^2 + (1/2)k * Xmax

I rearranged the equation for Xmax and got a final answer of:

**Xmax = (m1v1^2 + m2v2^2 - (m1 + m2)V^2)/k.**

c)

c)

From what I can tell, for this part of the question I am supposed to use the equation for elastic collisions in combination with the Conservation of Momentum equation to derive formulas for v1' and v2'.

For v1' I rearranged the equation m1v1 + m2v2 = m1v1' + m2v2' to isolate v2' and substituted for v2' in the equation v1 + v1' = v2 + v2' in order to make v1' the only unknown variable.

I ended up with the equation, v1' = v2 - v1 + (m1v1 + m2v2 - m1v1')/m2, and I rearranged this equation to get the answer

**v1' = v2(2m**. I followed the same logic to find v2'.

_{2}/(m1 + m2)) + v1((m1 - m2)/(m1 + m2)I'm reasonably sure that my logic would be sound if this were a horizontal surface. I cannot think of how the angle would contribute to the problem. Do I need to somehow incorporate mgh into into these equations?