Block hits block with spring. Conserv of Energy and Momentum?

• Fanman22
In summary, the conversation discusses a problem involving conservation of energy, specifically with a system of moving blocks and a compressed spring. The participants have difficulty finding the correct answer and discuss the concept of kinetic and potential energy, as well as linear momentum conservation. They also consider the effects of a frictionless surface and the role of inertia in the system. One participant ultimately finds the solution with the help of another.
Fanman22
Below is a copy of the problem and questions that are asked. I'm pretty sure that this has to be a conservation of energy problem, but I keep getting the wrong answer for the first problem so I can't continue and try the others.
I set the KE of the moving block equal to the PE of the spring that it strikes to determine X meters. I get 0.41m...which is wrong. Any help is greatly appreciated.

A 3.0 kg block slides along a frictionless tabletop at 7.0 m/s toward a second block (at rest) of mass 6.0 kg. A coil spring, which obeys Hooke's law and has spring constant k = 860 N/m, is attached to the second block in such a way that it will be compressed when struck by the moving block, Fig. 9-40.

What will be the maximum compression of the spring?
m
(b) What will be the final velocities of the blocks after the collision? (Assume the initial direction of the 3.0 kg block is positive.)
3.0 kg block
m/s
6.0 kg block
m/s

(c) Is the collision elastic? (Compare energy values to the nearest joule.)

yes***
no

Give the kinetic energy of the system before and after the collision to justify your answer.
before
J
after
J

The way i see it,linear momentum is conserved in the collision,but the total mechanical energy is conserved as well in 2 situations:globally and in the system spring-mass...

$m_{1}$ projectile's mass, $m_{2}$ target mass...

$$m_{1}\frac{\vec{v}_{1}^{2}}{2}=m_{1}\frac{\vec{v'}_{1}^{2}}{2}+m_{2}\frac{\vec{v'}_{2,MAX}^{2}}{2}$$

$$m_{1}\vec{v}_{1}=m_{1}\vec{v'}_{1}+m_{2}\vec{v'}_{2,MAX}$$

and of course

$$\frac{kA^{2}}{2}=\frac{m_{2}\vec{v'}_{2,MAX}^{2}}{2}$$

So i'd say that in the collision,the total KE is conserved,but after the bodies would no longer interact,in the next bilionth of the second,the velocity of tha target drops below the value acquired as a result of the collision,so the total balance is not valid anymore,till the body gets the same KE again,which is after T/2 and so on,because the target would oscillate forever with $T=2\pi \sqrt{\frac{m_{2}}{k}}$

So the last question is vague...

Daniel.

ok, basically you lost me after you said that linear momentum is conserved...but if that is the case, shouldn't the initial KE of the block be equal to the PE of the compressed spring to give me the distance that it compressed?

It would be the case,if the projectile had transferred all its KE in the collision.

U have to solve the system implied by the KE & momentum conservation in the collision

Daniel.

my equation is as follows:
.5(3)(7)^2 = .5(860)x^2
x=0.41m
...but its wrong.

This is strange...if the surface is frictionless, then should the spring even compress at all? Wouldn't ANY amount of force be directly transferred into velocity since there is no friction holding the second block in place?

Fanman22 said:
my equation is as follows:
.5(3)(7)^2 = .5(860)x^2
x=0.41m
...but its wrong.
It's wrong because you are setting the total energy equal to the spring potential energy. But even when the spring is maximally compressed, the system is moving so it has KE as well.

Hint: When the spring is at maximum compression, how does the speed of each mass compare to the speed of the center of mass of the system? (What's the speed of the center of mass?)

This is strange...if the surface is frictionless, then should the spring even compress at all? Wouldn't ANY amount of force be directly transferred into velocity since there is no friction holding the second block in place?
The spring is attached to a mass, which has inertia.

thanks, you helped me figure it out.

1. How does the conservation of energy apply to a block hitting a block with a spring?

The conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In this scenario, the energy from the first block hitting the spring is transferred to the second block, causing it to move.

2. What is the relationship between energy and momentum in this situation?

Momentum is the product of an object's mass and velocity, while energy is the ability to do work. In this scenario, when the first block hits the spring, it transfers some of its kinetic energy to the second block, causing it to gain momentum.

3. How does the spring affect the conservation of energy and momentum?

The spring acts as a medium for the transfer of energy and momentum between the two blocks. As the first block hits the spring, it compresses and stores potential energy. This potential energy is then transferred to the second block, causing it to gain kinetic energy and momentum.

4. Does the mass of the blocks affect the conservation of energy and momentum?

According to the law of conservation of energy and momentum, the total energy and momentum in a closed system remain constant. Therefore, the masses of the blocks do not affect the conservation of energy and momentum in this scenario.

5. What happens to the energy and momentum after the collision between the blocks?

After the collision, the first block loses some of its energy and momentum to the second block. The first block will also experience a change in its velocity, while the second block will continue to move with the momentum it gained from the collision. Overall, the total energy and momentum in the system remain constant.

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