Conservation of momentum/mass when dropping an object on a spring.

In summary: So in summary, the pan and steak collide with a negligible mass and force constant, and the steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. Immediately after the collision, the pan and steak have a speed of 2.56666666m/s.
  • #1
JustinLiang
81
0

Homework Statement


A spring of negligible mass and force constant k=400 is hung vertically and a 0.2kg pan is suspended from its lower end. A butcher drops a 2.2kg steak onto the pan from a height of 0.4m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What is the speed of the pan and steak immediately after the collision?

Homework Equations


PE=mgh
KE=mv^2/2
mv=mv
E(total energy in spring)=kA^2/2

The Attempt at a Solution


First I found out the energy produced from dropping the steak 0.4m:
PE=mgh
=2.2kg(9.8)(0.4m)
=8.624J

Then I used this to find the speed of the steak before the collision:
PE=KE
8.624J=(2.2kg)(v^2)/2
v=2.8m/s

When the steak goes into the pan by conservation of momentum we have:
mv=mv
(2.2kg)(2.8m/s)=(2.4kg)(v)
v=2.56666666m/s

My problem: However if I plug 2.566666m/s back into the KE equation I do not get 8.624J, instead I get 7.095J which means that the energy is not conserved? Do you guys know why? If I have 7.095J as the energy and I plug it into the E(total energy in spring) equation I do not get the right amplitude, instead plugging in 8.624J will get me the right answer.
HERE IS A SIMPLER PROBLEM
I found the same problem when solving for the final velocity of a block m=0.992kg when a bullet with v=280m/s and m=0.008kg strikes it and embeds itself into it.

By conservation of momentum:
(0.008)(280)=(1kg)(v)
v=2.24m/s

Where initially the KE in the bullet initially is
KE=0.008(280)^2/2=313.6J

But the KE after the collission is
KE=1kg(2.25m/s)^2/2=2.5088J

So clearly the energy is not conserved? I am confused.
 
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  • #2
JustinLiang said:
When the steak goes into the pan by conservation of momentum we have:
mv=mv
(2.2kg)(2.8m/s)=(2.4kg)(v)
v=2.56666666m/s

My problem: However if I plug 2.566666m/s back into the KE equation I do not get 8.624J, instead I get 7.095J which means that the energy is not conserved? Do you guys know why? If I have 7.095J as the energy and I plug it into the E(total energy in spring) equation I do not get the right amplitude, instead plugging in 8.624J will get me the right answer.

You have done it correctly, and it is the correct answer. So you are right that energy is not conserved during the collision. That's why the collision is called inelastic.
 
  • #3
Any energy 'lost' has just been converted into heat and sound during the collision :)
 
  • #4
BruceW said:
You have done it correctly, and it is the correct answer. So you are right that energy is not conserved during the collision. That's why the collision is called inelastic.

If that's the case shouldn't 7J get me the correct amplitude? The answer suggests I use 8.6J, but this energy is not conserved?
 
  • #5
7J should get you the right amplitude. Maybe the answer book is incorrect?
 
  • #6
hmm, actually maybe the lost KE is turned into PE in the spring (so that although the collision is fully inelastic, energy is conserved).
 
  • #7
BruceW said:
hmm, actually maybe the lost KE is turned into PE in the spring (so that although the collision is fully inelastic, energy is conserved).

Yeah it is totally inelastic so all the energy would be conserved. But how do you explain the differences in the energy before collision 8.624J and after collision 7.905J (I initially typed out the wrong values).
 
  • #8
I just tried a different calculation:

If all the energy is conserved then:
PE=KE(with both masses)
8.624J=(2.4kg)(v^2)/2
v=2.68m/s

This answer tells us all the PE becomes KE in the combined mass. However the answer key has a v=2.6m/s so I don't think this is right. Maybe the answer key is wrong...
 
  • #9
I'm guessing the answer key is wrong. It does happen on occasion.
 
  • #10
Okay thanks!
 

1. What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant in the absence of external forces.

2. How does the conservation of momentum apply to dropping an object on a spring?

When an object is dropped on a spring, the object's momentum is transferred to the spring, causing it to compress. The total momentum of the system (object + spring) remains constant, but is distributed between the two objects.

3. What is the relationship between mass and momentum in the conservation of momentum?

According to the conservation of momentum, the total momentum of a system is equal to the sum of the momenta of all the objects within the system. This means that the mass of an object is directly proportional to its momentum.

4. How does the conservation of mass apply to dropping an object on a spring?

The conservation of mass is a principle in physics that states that matter cannot be created or destroyed, only transformed. When an object is dropped on a spring, the mass of the object remains constant, but its potential energy is converted to elastic potential energy in the spring.

5. What are some real-world applications of the conservation of momentum/mass when dropping an object on a spring?

The conservation of momentum/mass when dropping an object on a spring is seen in everyday objects such as pogo sticks, trampolines, and shock absorbers in vehicles. It is also applied in engineering and design of structures to absorb impact and reduce damage. In addition, it is a key concept in understanding collisions and explosions in physics.

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