How Far Do Colliding Blocks Slide on a Spring?

In summary, the conversation discusses a scenario where a block attached to a spring collides with another block of twice its mass. The question is how far the blocks will slide beyond the collision point. The solution uses conservation of energy and Newton's second law to calculate the velocity and acceleration of the blocks at the point of collision. The final answer is that the blocks will slide a distance of 3/8 times the initial compression distance of the spring.
  • #1
naianator
48
1

Homework Statement



problems_MIT_boriskor_BKimages_10-mass-spring-two-block-collision.png


A block of mass m is attached to a spring with a force constant k, as in the above diagram. Initially, the spring is compressed a distance x from the equilibrium and the block is held at rest. Another block, of mass 2m, is placed a distance x/2 from the equilibrium as shown. After the spring is released, the blocks collide inelastically and slide together. How far (Δx) would the blocks slide beyond the collision point? Neglect friction between the blocks and the horizontal surface.

Homework Equations


U_spring = 1/2*k*x^2

K = 1/2*m*v^2

F = m*a

v_f^2 = v_0^2+2*a*x

The Attempt at a Solution


E_i = 1/2*k*x^2 = E_f = 1/2*m*v^2 + 1/2*k*x_f^2

at x/2 where m collides with 2m:

1/2*k*x^2 = 1/2*3m*v^2 + 1/8*k*x^2

k*x^2 = 3m*v^2 + 1/4*k*x^2

3/4*k*x^2 = 3*m*v^2

yielding

v = sqrt(k*x^2/(4*m))

by Newtons second law a = -k*x/(3*m)

and using kinematics (v_f^2 = v_0^2+2*a*x)

0 = k*x^2/(4*m) - 2*(k*x/(3*m))*delta(x)

x/4 = 2/3*delta(x)

delta(x) = 3/8*x

Am I way off base with this approach?
 
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  • #2
Could you possibly explain what you have written , in words ?
It is a bit difficult to understand what you have typed .
 
  • #3
Qwertywerty said:
Could you possibly explain what you have written , in words ?
It is a bit difficult to understand what you have typed .
I used conservation of energy to find the velocity where the two blocks collide at x/2. I'm not sure about this though, because I don't think that conservation of momentum applies and I used 3m for the mass (assuming that the blocks had collided). Then I used N2L to find the acceleration based on the force (-kx) and mass (3m). I plugged that into the kinematics equation for final velocity and solved for delta x.
 
  • #6
naianator said:
Yes, sorry, I wasn't having any luck with responses and it was due a half hour ago so I reposted it. Is there a way to delete this thread?
No, but you might be able to edit the title to something that will stop readers wasting their time delving into it.
A moderator could delete it.
 

1. How does the spring affect the collision between the block and the surface?

The spring affects the collision by providing a force that opposes the motion of the block, causing it to compress and then expand back to its original position. This results in a change in the momentum of the block and the force exerted on the surface.

2. What factors affect the magnitude of the force exerted by the spring during the collision?

The magnitude of the force exerted by the spring depends on the stiffness of the spring, the mass and velocity of the block, and the duration of the collision. A stiffer spring, higher velocity, and longer duration of the collision will result in a larger force exerted by the spring.

3. How does the mass of the block affect the collision with the spring?

The mass of the block affects the collision by determining the amount of kinetic energy that is transferred to the spring during the collision. A heavier block will have more kinetic energy and therefore transfer more energy to the spring, resulting in a larger compression and expansion of the spring.

4. Can the spring be compressed beyond its elastic limit during the collision?

Yes, if the force exerted by the block on the spring exceeds the elastic limit of the spring, it can be permanently deformed. This can result in a decrease in the force exerted by the spring during subsequent collisions.

5. What is the relationship between the frequency of the spring and the frequency of the block-spring system after the collision?

The frequency of the spring and the frequency of the block-spring system after the collision are directly related. As the block compresses and expands the spring, it also causes it to vibrate at a certain frequency. The frequency of the system will be the same as the frequency of the spring if the mass of the block is negligible compared to the mass of the spring.

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