Problem about an impulse and a spring

In summary: Yes. If the upper block moves faster, the string will be compressing, if it is moving slower it will be decompressing. This should give you enough information to solve your original question.
  • #1
ubergewehr273
142
5

Homework Statement


The attached image.

Homework Equations


Momentum conservation
Conservation of mechanical energy

The Attempt at a Solution


I tried conserving momentum,
##P=mv_{1} + Mv_{2}##
and then conserving M.E.,
P^2/(2m)=1/2##mv_1{1}^2## + 1/2##Mv_{2}^2##
After that I can't seem to relate v1 and v2. Also can someone tell me the condition when maximum compression of spring is obtained ?
 

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  • #2
ubergewehr273 said:
Also can someone tell me the condition when maximum compression of spring is obtained ?
What do you think? Can you think of an expression for how fast the compression of the spring is growing?
 
  • #3
Orodruin said:
What do you think? Can you think of an expression for how fast the compression of the spring is growing?
I think when mass m stops moving relative to the ground is when max compression occurs.
 
  • #4
ubergewehr273 said:
I think when mass m stops moving relative to the ground is when max compression occurs.
This is not correct. Can you give an expression for the spring length as a function of the positions of the masses?
 
  • #5
Orodruin said:
This is not correct. Can you give an expression for the spring length as a function of the positions of the masses?
If mass m moves by x1 wrt mass M and mass M moves by x2 wrt ground then spring compression will be x1-x2.
 
  • #6
Right, so what is the time derivative of that expression? That would be the change in the compression per time. What does this tell you about when the compression is maximal?
 
  • #7
Orodruin said:
Right, so what is the time derivative of that expression? That would be the change in the compression per time. What does this tell you about when the compression is maximal?
So compression is maximum when both masses have equal velocities.
 
  • #8
ubergewehr273 said:
So compression is maximum when both masses have equal velocities.
Yes. If the upper block moves faster, the string will be compressing, if it is moving slower it will be decompressing. This should give you enough information to solve your original question.
 

1. What is an impulse in relation to a spring?

An impulse is a force acting over a short period of time, resulting in a change in momentum. In the case of a spring, an impulse can cause the spring to compress or extend, depending on the direction of the force.

2. How does an impulse affect the motion of a spring?

An impulse can affect the motion of a spring by changing its velocity and/or position. If the impulse is in the same direction as the current motion of the spring, it will increase the velocity and extend the spring further. If the impulse is in the opposite direction, it will decrease the velocity and compress the spring.

3. What factors affect the magnitude of an impulse on a spring?

The magnitude of an impulse on a spring is affected by the force applied and the duration of the force. A greater force or a longer duration will result in a larger impulse. The mass of the spring and any external factors, such as friction, may also affect the magnitude of the impulse.

4. Can the elasticity of a spring affect the impulse?

Yes, the elasticity of a spring can affect the impulse. A more elastic spring will be able to absorb and release a greater amount of energy, resulting in a larger impulse. On the other hand, a less elastic spring may not be able to handle as much energy and will have a smaller impulse.

5. How can the motion of a spring be modeled using the impulse-momentum theorem?

The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse acting on it. This can be applied to the motion of a spring by considering the initial and final momentum of the spring after an impulse is applied. By equating these two values and solving for the unknown variables, the motion of the spring can be accurately modeled.

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