Dropping a Block on a Spring with Completely Inelastic Collision

In summary, the conversation discusses a problem involving a block of mass m dropped onto a plate of mass M attached to a spring with spring constant k. The system starts to oscillate and the question is posed about the amplitude of the oscillations. The conversation delves into an approach using conservation of energy and discusses the conversion of gravitational potential energy to kinetic energy during the initial drop, as well as the conversion of kinetic energy to elastic potential energy during the oscillations. The conversation also addresses a previously made mistake in calculating the potential energy converted during the oscillations and clarifies that both the block and plate move downwards. The final conclusion is that the potential energy converted during the oscillations is (m+M)gx.
  • #1
SeventeenForever

Homework Statement

[/B]
A block of mass, m, is dropped from height, h (above the plate), onto a plate of mass, M, which is attached to a spring with spring constant, k. The block sticks to the plate and the system starts to oscillate. What is the amplitude of the oscillations.

Homework Equations


Etotal=K + Ugravitational + Uspring
Uspring=0.5kx2
K=0.5mv2
Ugravitational=mgh
Conservation of Energy

The Attempt at a Solution



I decided to approach this problem from an energy perspective, setting the equilibrium position of the spring (w/ plate) as 0 gravitational potential energy.

The initial gravitational potential energy is Ug=mgh. Just before it hits the plate, all of this gravitational PE is converted to KE.

mgh = 0.5mv2

v= (2gh)0.5

The problem specifically states that the block "sticks" to the plate, suggesting a completely inelastic collision. I can use the conservation of momentum to find the velocity after the collision; after the collision, the velocity of the plate and block is:

vf= m(2gh)0.5 / (m+M)

The kinetic energy of the system is now:

0.5(m+M)vf2 = Kf

The spring will then be compressed from its equilibrium position until the velocity of the plate/block becomes zero, which is the max compression/amplitude. At this point, all of the kinetic energy has been converted to elastic potential energy.

0.5(m+M)vf2 = 0.5kx2

I can then plug in my equation for vf and solve for x, which is the maximum compression/amplitude.

Is this approach okay? Did I make any incorrect assumptions?

Thaaaaaanks!
 
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  • #2
SeventeenForever said:
At this point, all of the kinetic energy has been converted to elastic potential energy.
Did gravity mysterously turn off at point of impact?
 
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  • #3
I didn't think about that since I set my reference point to be the equilibrium position of the spring.

So after collision, the gravitational potential energy from the height, h, was either converted to kinetic energy or lost in the inelastic collision. The spring will compress until there is no kinetic energy, but some of this energy will be stored up in gravitational potential energy?

0.5(m+M)vf2=(m+M)gx + 0.5kx2 + 0.5(m+M)(0)2

This would then be a quadratic formula and I could solve for x.

I'm a little hesitant on this approach now since I appear to be gaining gravitational potential energy as the plate/block goes downward. Is this approach at all possible or should I just restart and set the ground as potential energy zero and work from there?

Thanks for the reply!
 
  • #4
SeventeenForever said:
some of this energy will be stored up in gravitational potential energy?
Er.. no.
As the spring compresses the masses descend. So what is happening to GPE?
 
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  • #5
It would continue to decrease normally. Would it be converted to elastic potential energy as well? I'm getting tripped up since I defined the top of the spring in equilibrium position to be 0 GPE. I suppose it would also decrease in this case too since the height is decreasing and that's what gravitational potential energy does.
 
  • #6
Correct. The total height for the purpose of calculating GPE is h plus however much the spring compresses.
 
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  • #7
So to recap, the kinetic energy that remains is converted to elastic PE. As the masses descend on top of the spring, their potential energies are converted to elastic potential energies as well:

m2gh/(m+M) + max = 0.5kx2
(the m2gh/(m+M) is a simplified kinetic energy term)

This is a quadratic equation I can solve.
 
  • #8
SeventeenForever said:
So to recap, the kinetic energy that remains is converted to elastic PE. As the masses descend on top of the spring, their potential energies are converted to elastic potential energies as well:

m2gh/(m+M) + max = 0.5kx2
(the m2gh/(m+M) is a simplified kinetic energy term)

This is a quadratic equation I can solve.
Yes (where max=?).
 
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  • #9
My bad; I made a typo. Max should be mgx (autocorrect)
 
  • #10
SeventeenForever said:
Max should be mgx
So, the block is passing through the plate?
 
  • #11
Oh the block and the plate both move downwards. So the amount of potential energy converted is actually (m+M)gh. Thanks for catching that!
 
  • #12
SeventeenForever said:
Oh the block and the plate both move downwards. So the amount of potential energy converted is actually (m+M)gh. Thanks for catching that!
(m+M)gx.
 
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Related to Dropping a Block on a Spring with Completely Inelastic Collision

1. What is a completely inelastic collision?

A completely inelastic collision is a type of collision where two objects collide and stick together after impact. In this type of collision, kinetic energy is not conserved and some energy is lost in the form of heat or sound.

2. How does dropping a block on a spring with a completely inelastic collision work?

When a block is dropped onto a spring, it compresses the spring due to the force of gravity. The block and the spring then collide with each other, and if the collision is completely inelastic, the block will stick to the spring and continue to move with it as it decompresses.

3. What factors affect the outcome of dropping a block on a spring with a completely inelastic collision?

The outcome of this experiment can be affected by factors such as the mass and velocity of the block, the stiffness of the spring, and the height from which the block is dropped. These factors can impact the amount of energy lost during the collision and the resulting motion of the block and spring system.

4. How is energy conserved in a completely inelastic collision?

In a completely inelastic collision, kinetic energy is not conserved as some energy is lost in the form of heat or sound. However, the total energy of the system (including potential and kinetic energy) is conserved. This means that the energy lost during the collision is converted into other forms of energy within the system.

5. What are some real-world applications of studying dropping a block on a spring with a completely inelastic collision?

This experiment can be used to study the concept of impulse and momentum, which is important in understanding the behavior of objects in collisions. It can also be applied to real-life situations such as car accidents or sports collisions, where understanding the transfer of energy and momentum is crucial in predicting outcomes and preventing injuries.

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