Kinetic energy and collision involving a spring

In summary, the question asks which mass, when given 16J of kinetic energy, will result in the greatest compression of the spring. The answer is the 2-kg mass, as it has a smaller mass and thus a larger change in momentum, leading to a larger force and greater compression of the spring. This can be proven by considering the conservation of momentum and using the equation F=dp/dt.
  • #1
NeuroFuzzy
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I'm not quite happy with my justification of a problem. I'm not posting this in the homework section because it's not homework, and my trouble with doesn't revolve around finding the answer. The problem is this:
A 2-kg mass and an 8-kg mass collide elastically, compressing a spring bumper on one of them.
part 1) Which was of causing the collision to occur will result in the greater compression of the spring: Giving the 8-kg mass 16J of kinetic energy or giving the 2-kg mass 16J of kinetic energy? (the other mass remains stationary.
part 2) keeping the total kinetic energy to be 16J, how should this energy be divided to obtain the greatest compression of the spring?

I based my answer around kinetic energy. We can let [itex]v_1[/itex] and [itex]v_2[/itex] be the velocities of the masses relative to the center of mass, and [itex]v[/itex] be the velocity of the center of mass. The total initial energy of the system (before the spring is compressed) as: [itex]E=\frac{m_1}{2}v_1^2+\frac{m_1}{2}v_1^2+\frac{M}{2}v^2[/itex]. (where [itex]M[/itex] is the total mass of the system) My argument is that we want to minimize the amount of energy locked up in the movement of the center of mass so that the energy can be used to give potential energy to the spring. So then:
[itex]\frac{M}{2}v^2=\frac{m_1+m_2}{2}\cdot \frac{(m_1 v_1+m_2 v_2)^2}{(m_1+m_2)^2}[/itex].
We also have [itex]J=\frac{1}{2}mv^2\rightarrow v=\sqrt{\frac{2 J}{m}}[/itex]. The other velocity is zero, so:
[itex]\frac{(m_1 v_1+m_2 v_2)^2}{2(m_1+m_2)}=\frac{(m_1 v_1)^2}{2(m_1+m_2)}[/itex]
Since the bottom is constant we can just consider the top term
[itex]=m_1^2 v_1^2=m_1^2\frac{2 J}{m_1}=m_1 2 J[/itex]
To minimize this term, we obviously need to use the 2kg mass. The second part can be completed in the same way by setting the center-of-mass's velocity to be zero.

My question is... how can I justify this better? I'm really uncomfortable with energy based arguments
 
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  • #2
I do not quite see why did you use the idea with the mass center. Besides, I think you have doubled the energy in your term for [itex]E[/itex], but I might have misunderstood something.

Anyway, why don't you simple state, that the larger compression will be the result of a larger force in the spring. And since [itex]F = dp/dt[/itex], then the larger change of momentum will imply larger force.

Now, [itex]dp = p_2 - p_1[/itex] and if we attach the spring to the mass that was initialy at rest, we have [itex]p_1 = 0[/itex]. So [itex]F \propto p_2[/itex]. You will now need to consider two cases:

1. Moving mass [itex]m_2=8 kg[/itex]
2. Moving mass [itex]m_2 = 2 kg[/itex]

From each of them you will receive different velocities of this mass. Then you apply the conservation of momentum to find the velocity [itex] v_f [/itex] of the whole system (moving together as one body) after the impact. Having this velocity you will be able to establish [itex] p_2[/itex] for each case and compare them.
 

FAQ: Kinetic energy and collision involving a spring

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is a scalar quantity and is dependent on the mass and velocity of an object.

2. How is kinetic energy related to collision involving a spring?

In a collision involving a spring, the kinetic energy of the object before the collision is converted into potential energy of the spring as it is compressed, and then back into kinetic energy as the spring pushes the object away. This transfer of energy is what allows the object to continue moving after the collision.

3. How does the spring constant affect the kinetic energy in a collision involving a spring?

The spring constant, which is a measure of the stiffness of the spring, affects the amount of potential energy stored in the spring during compression. This, in turn, affects the amount of kinetic energy that can be transferred to the object during the collision.

4. Is kinetic energy conserved in a collision involving a spring?

In an ideal scenario, where there is no external force acting on the system, the total kinetic energy of the system (object and spring) is conserved. However, in real-world situations, some energy may be lost due to factors such as friction.

5. How can the kinetic energy in a collision involving a spring be calculated?

The kinetic energy in a collision involving a spring can be calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity. Alternatively, it can also be calculated by subtracting the potential energy of the spring from the total energy of the system after the collision.

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