# How do I solve for an elastic collision going up a ramp?

• Temple1998
In summary: Would it be incorrect to use the equation 1/2m1v1^2 + mgh1 + 1/2m2v2^2 +mgh2 = 1/2(m1 + m2)V'^2 +(m1 + m2)gh'? And I'm assuming I would use xsin30 for the heights?Yes, and the equation would also need to take into account the accelerations of the blocks. No, the equation is correct.In summary, the two blocks with masses m1=3kg and m2=4.5kg move on a platform with velocities v1 and v2, respectively. The platform is inclined with θ
Temple1998

## 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:
1. right before touches the spring where the two objects move with v1 and v2 (This picture shows the blocks at positions A and B).
2. the spring is at maximum compression and the system moves with common V (This picture shows the blocks at positions C and D).
3. the two blocks lost contact and move with separate speeds (This picture shows the blocks at positions E and F).
a) Find the velocity V at maximum compression,
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)

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)

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(2m2/(m1 + m2)) + v1((m1 - m2)/(m1 + m2). I followed the same logic to find v2'.

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?

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Temple1998 said:
I used the Conservation of Momentum equation: m1v1 + m2v2 = (m1 + m2)V
What are the conditions necessary for conservation of momentum in a given direction in a given subsystem? Are those conditions satisfied for the subsystem consisting of the two blocks, in a direction parallel to the ramp, for the duration of the transition from state 1 to state 2?

haruspex said:
What are the conditions necessary for conservation of momentum in a given direction in a given subsystem? Are those conditions satisfied for the subsystem consisting of the two blocks, in a direction parallel to the ramp, for the duration of the transition from state 1 to state 2?
Would gravity cause a net external force due to its x component? I don't know how to go about this problem if this is the case and I can't use the conservation of momentum equation. I just learned conservation of momentum and I haven't been able to find any other problems like this to reference.

Temple1998 said:
Would gravity cause a net external force due to its x component?
Yes, and its contribution to momentum will depend on the elapsed time. Therefore you need to find the SHM equation describing the motion from state 1 to state 2 (unless you can think of a way of jumping straight to the oscillation period from the given constants).

haruspex said:
Yes, and its contribution to momentum will depend on the elapsed time. Therefore you need to find the SHM equation describing the motion from state 1 to state 2 (unless you can think of a way of jumping straight to the oscillation period from the given constants).
Would it be incorrect to use the equation 1/2m1v1^2 + mgh1 + 1/2m2v2^2 +mgh2 = 1/2(m1 + m2)V'^2 +(m1 + m2)gh'? And I'm assuming I would use xsin30 for the heights? I don't have any experience with simple harmonic motion in relation to two objects.

Temple1998 said:
Would it be incorrect to use the equation 1/2m1v1^2 + mgh1 + 1/2m2v2^2 +mgh2 = 1/2(m1 + m2)V'^2 +(m1 + m2)gh'? And I'm assuming I would use xsin30 for the heights? I don't have any experience with simple harmonic motion in relation to two objects.
You can certainly use conservation of energy, but you need to consider the energy change in the spring. I'm not sure whether that will give you enough equations, though.

haruspex said:
You can certainly use conservation of energy, but you need to consider the energy change in the spring. I'm not sure whether that will give you enough equations, though.
I'm stumped at this point, then.

Temple1998 said:
I'm stumped at this point, then.
It can help to think in terms of the mass centre as far as possible, treating the oscillation as superimposed on the movement of that.
What will be the acceleration along the slope of the mass centre? What is its initial velocity?
Can you write an equation relating those to the given distances and to the time elapsed from state 1 to state 3?

Are v1 and v2 4ms-1 and 2ms-1 respectively - can't quite make out from the diagram of slope and blocks at the top ?

Since the influence of g is the same for both blocks, I am wondering if one can assume momentum conservation within the 'block frame' , solve the collision problem using standard methods, and then somehow translate the answers back to 'earth frame' ?

Temple1998 said:
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). seems fine

b)

I used the equation: (1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)V^2 + (1/2)k * Xmax shouldn't there be a square here ?

I rearranged the equation for Xmax and got a final answer of: Xmax = (m1v1^2 + m2v2^2 - (m1 + m2)V^2)/k.

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(2m2/(m1 + m2)) + v1((m1 - m2)/(m1 + m2). I followed the same logic to find v2'. all good for collision on a flat plane

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? I think we just need to 'translate' the velocities determined above using something like v' = v - gsin(30)Δt. T(spring) = https://socratic.org/questions/frequency-of-vibration-of-two-masses-connected-by-a-spring

Last edited:

## 1. How do I determine the initial velocity of the object?

The initial velocity of the object can be determined by using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. You can also use the conservation of energy equation to calculate the initial velocity.

## 2. What is the formula for calculating the final velocity of the object?

The formula for calculating the final velocity of the object in an elastic collision going up a ramp is v = √(u^2 + 2gh), where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and h is the height of the ramp.

## 3. How do I determine the angle of the ramp in an elastic collision?

The angle of the ramp can be determined by using the inverse tangent function (tan^-1). You will need to know the height and horizontal distance of the ramp in order to calculate the angle.

## 4. Is there a specific method for solving elastic collisions going up a ramp?

Yes, there is a specific method for solving elastic collisions going up a ramp. First, you will need to calculate the initial velocity and angle of the ramp. Then, you can use the conservation of energy equation to find the final velocity of the object. Finally, you can use the formula for calculating the final velocity to determine the speed and direction of the object after the collision.

## 5. Can I use the same equations for both elastic and inelastic collisions on a ramp?

No, the equations for elastic and inelastic collisions are different. In an inelastic collision, some of the kinetic energy is lost due to friction and heat, so the equations will be different. However, in both types of collisions, the conservation of momentum will still apply.

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