Kinetic energy and collision involving a spring

AI Thread Summary
The discussion centers on the dynamics of an elastic collision involving a 2-kg and an 8-kg mass compressing a spring. The first part of the problem examines which mass, when given 16J of kinetic energy, results in greater spring compression, with the conclusion that the 2-kg mass is preferable due to its ability to minimize energy in the center of mass motion. The second part seeks to determine the optimal division of 16J of total kinetic energy for maximum spring compression, suggesting that the center-of-mass velocity should be zero to maximize potential energy transfer to the spring. Concerns are raised about the justification of energy-based arguments and the potential doubling of energy in calculations. The discussion concludes with a recommendation to analyze momentum changes to better understand the relationship between force and spring compression.
NeuroFuzzy
Messages
8
Reaction score
0
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 v_1 and v_2 be the velocities of the masses relative to the center of mass, and v be the velocity of the center of mass. The total initial energy of the system (before the spring is compressed) as: E=\frac{m_1}{2}v_1^2+\frac{m_1}{2}v_1^2+\frac{M}{2}v^2. (where M 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:
\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}.
We also have J=\frac{1}{2}mv^2\rightarrow v=\sqrt{\frac{2 J}{m}}. The other velocity is zero, so:
\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)}
Since the bottom is constant we can just consider the top term
=m_1^2 v_1^2=m_1^2\frac{2 J}{m_1}=m_1 2 J
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
 
Physics news on Phys.org
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 E, 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 F = dp/dt, then the larger change of momentum will imply larger force.

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

1. Moving mass m_2=8 kg
2. Moving mass m_2 = 2 kg

From each of them you will receive different velocities of this mass. Then you apply the conservation of momentum to find the velocity v_f of the whole system (moving together as one body) after the impact. Having this velocity you will be able to establish p_2 for each case and compare them.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top