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A ball of mass $m$ is attached to a spring that is attached to a heavy moving cart (mass of cart $M >>$ mass of ball $m$) moving at velocity $V0$. The compressed spring with known spring constant $k$ is released and the ball is fired forward. The ball then strikes a similar spring with the same spring constant attached to a wall (i.e. the spring is attached to a heavy immovable object). How far does the spring attached to the wall compress?

Is it just a matter of conservation of energy? Ek(moving ball) + E(spring potential-for the moving spring) = E(spring potential-fixed spring)

Or does the motion of the initial spring somehow affect how much energy it can transmit to the ball.

Or, in the frame of reference of the moving spring, can you simply calculate the velocity of the ball from E(spring potential) = Ek and add that to the initial velocity of the spring/ball and then use that new velocity to calculate the Ek and hence the energy transferred to the fixed spring and hence the compression of the spring.

Depending on how you approach this problem you get different answers.

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  • $\begingroup$ If you're using the frame of the moving spring, then the "fixed spring" is not at rest. So the spring must compress sufficiently not to stop the ball, but to bring the ball to the same velocity as the wall. $\endgroup$ – BowlOfRed Dec 15 '18 at 7:53
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Or does the motion of the initial spring somehow affect how much energy it can transmit to the ball.

The motion of the heavy cart must also be included in the calculation. During the launch of the ball the heavy cart is both moving and exerting a force in the direction of motion on the spring therefore work is done on the spring. So the force adds energy to the spring which is then transferred to the ball. So the total mechanical energy delivered to the receiving spring increases with increasing speed of the heavy cart. The formula that you wrote for energy is on the right track, but incomplete.

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  • $\begingroup$ cart moves at constant velocity here. "force" doesn't seem appropriate to me? $\endgroup$ – QuIcKmAtHs Dec 15 '18 at 5:27
  • $\begingroup$ Where in Newton’s three laws does it say that you can neglect a force if something moves at constant velocity? On a free body diagram you include all forces regardless of the motion of the source. $\endgroup$ – Dale Dec 15 '18 at 13:02
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Of course, I assume no drag forces or frictional forces are involved here. Initially, while the cart in moving at a velocity $v$, the ball possesses energy of $\frac{1}{2}mv^2$.

Along with the elastic potential of the energy of $\frac{1}{2}kx_1^2$, the sum of this energy will be converted to the kinetic energy of the moving ball, which is then converted to elastic potential energy of the second spring.

The equation should be as follows: $\frac{1}{2}mv^2+\frac{1}{2}kx_1^2=\frac{1}{2}kx_2^2$

However, it is flawed to calculate the amount of velocity the spring would impart to the ball and then summing it up with the cart's velocity (also the ball's velocity). This method can only work if the ball is initially stationary, meaning, the ball is stationary too. Energy is proportional to $v^2$, to increase each an object by $1ms^{-1}$, it requires more energy each time. Thus, we cannot do it that way.

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  • $\begingroup$ Thank you for the quick feedback - I agree and this is how I would solve it, however, I was asked why you could not use the spring potential to calculate the "launch" velocity of the ball and then add this velocity to the original velocity to get the final velocity of the ball . Then this final velocity would be used as 1/2mv^2 = 1/2kx2^2. $\endgroup$ – gmarlow Dec 15 '18 at 3:55
  • $\begingroup$ @gmarlow I will reply to that shortly, I understand now. $\endgroup$ – QuIcKmAtHs Dec 15 '18 at 3:57
  • $\begingroup$ You are missing the work done by the cart $\endgroup$ – Dale Dec 15 '18 at 4:52
  • $\begingroup$ @Dale the cart is moving at constant velocity $\endgroup$ – QuIcKmAtHs Dec 15 '18 at 5:26
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In Newtonian mechanics, the total mechanical energy is not frame-independent. If you calculate the energy of the ball/cart/spring/spring system in the rest frame of the walls of the room you get one value. If you calculate it in the rest frame of the cart you get a different value. That means you can't directly use the energy calculated in the cart frame as the value in the walls-of-room frame.

If you want to use the mechanical energy of the cart frame, you have to realize that the spring attached to the wall is moving toward the ball which affects the maximum compression of the spring.

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A thought experiment: you’re in a train moving at 10 m/s, and inside it there’s a 1kg object moving forward at 10 m/s with respect to the train. Inside the train, you calculate its kinetic energy to be 50 J. When it was at rest with respect to the train, an outside observer would also have given it 50 J. But now, the outside observer calculates its energy to be 200 J (not 50+50). You can’t add up energy like that.

A spring at rest imparts an energy of $\frac12 k x^2$ to an object, which is calculated by integrating the force over a sequence of infinitesimal distances. But when moving, the spring exerts those same forces over longer distances, so it does more work and imparts more energy. Again, adding up energy from the ‘inertial frame’ and from whatever happens at rest within that frame leads to the wrong result.

Thus, your intuition was wrong: you have to add up the velocities, not the energies. You may ask where the extra energy comes from (both in your experiment and in the one in the train). It comes from the vehicle, which exerts a force on one side of the spring and thus does work. If it has finite mass, it will slow down a bit and donate some of its kinetic energy to the object. If its mass is infinite, its kinetic energy also goes down by that amount – from infinity to infinity.

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  • $\begingroup$ And this gets downvoted because... ? $\endgroup$ – Jim Danner Dec 15 '18 at 9:35
  • $\begingroup$ To further the train thought experiment…. If on the train, the spring constant was 100N/m and the spring was compressed 1m, it would impart 50J of energy to the ball (in the train's frame of reference). So the ball would be moving 10 m/s now relative to the train. Would it not be accurate to say the ball is moving at 20m/s relative to an outside observer and thus has 200J of energy? So from the frame outside the train, the change in energy was +150J caused by the moving spring imparting a force on the ball. $\endgroup$ – gmarlow Dec 15 '18 at 13:41
  • $\begingroup$ @gmarlow Yes, that is accurate. It shows that we can’t simply add up those energies. $\endgroup$ – Jim Danner Dec 15 '18 at 17:26
  • $\begingroup$ Hey Jim - thanks for the comments - I would have upvoted your comment but since i am a newbie, my upvotes don't count yet.... $\endgroup$ – gmarlow Dec 15 '18 at 17:34
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If you want to use conservation of Mechanical Energy, use instead work energy theorem because conservation principles are frame dependent, but work energy theorem is frame independent.

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