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In my AP Physics C class today, we ran into a problem written by College Board whose answer we disputed. The problem is as such :

A block of mass $M=5.0 \ \mathrm{kg}$ is hanging at equilibrium from an ideal spring with spring constant $k=250 \ \mathrm{N/m}$.An identical block is launched up into the first block. The new block is moving with a speed of $v=5.0 \ \mathrm{m/s}$ when it collides with and sticks to the original block. Calculate the maximum compression of the spring after the collision of the two blocks.

According to the College Board answer key, the answer is $0.5 \ \mathrm{m}$ :

$p_1=p_2$

$Mv_0=(M+M)v_2$

$v_2=\frac{1}{2}v_0= \left (\frac{1}{2} \right)\left (5.0 \frac{m}{s}\right)$

$v_2=2.5 \frac{m}{s}$

$K_1 + U_1=K_2+U_2$

$\frac{1}{2}mv_1^2 +0=0+\frac{1}{2}kx_2^2$

$x_2=\sqrt{\frac{m}{k}}v_1= \sqrt{\frac{(10 \ \mathrm{kg)}}{\left(250 \frac{N}{m}\right)}} \left(2.5 \frac{m}{n}\right)$

$x_2=0.50 \ \mathrm{m}=50 \ \mathrm{cm}$

However, half of us disputed this during class. We argued that, yes, $U_2$ includes $\frac{1}{2}kx^2$, but it also includes gravitational potential energy at the maximum compression (that is, when it compresses $x$ meters from equilibrium, the mass $M$ is $x$ meters higher above ground). Thus $K_1+U_1=K_2+U_2$ is $\frac{1}{2}mv^2+0=0+\frac{1}{2}kx^2+mgx$. When $mgx$ is included, $x$ is $0.24 \ \mathrm{m}$, not $0.5 \ \mathrm{m}$.

My physics teacher reluctantly agreed with College Board but could not give a solid explanation why. He said he would e-mail College Board, but in the meantime, I would very much appreciate any input from people who know the answer.

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  • $\begingroup$ I agree with your approach. It should be 2mgh since the mass rising is two blocks of 5 kg each. $\endgroup$
    – Dan
    Dec 1, 2021 at 19:56
  • $\begingroup$ Sorry I should have written that lol. When I plugged the numbers in and solved, I did make sure to plug in 10 kg (times 9.81 m/s/s times x). $\endgroup$ Dec 1, 2021 at 19:57
  • $\begingroup$ Someone in class pointed out that perhaps because the question asks for MAXIMUM compression, they assume gravity does not exist. However, this still doesn't make sense. Assuming gravity does not exist (perhaps it is laid horizontally), the spring would have no elongation. When it is laid vertically, it elongates until the tension of the coil matches Mg. Thus, its equilibrium when horizontal (or simply without gravity) will be the length of the fully compressed spring. If the second block struck it elastically, the spring could not compress anymore. A compression of .5 meters is improbable. $\endgroup$ Dec 1, 2021 at 20:08
  • $\begingroup$ Hello! It is preferable to type out screenshots or images of text; for formulae, one can use MathJax. Thanks! $\endgroup$
    – jng224
    Dec 1, 2021 at 20:24
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    $\begingroup$ Don't forget to include the initial elongation of the spring. $\endgroup$
    – DanDan面
    Dec 1, 2021 at 20:36

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I guess this is a homework/check-my-work problem, so by the letter of the law I should not answer, but I would argue there is broad interest in solving it correctly given that a supposedly reputable source is presenting an incorrect solution.


Here is how I would do this. Initially, the spring is stretched distance $d=mg/k$ below its equilibrium position. Choose the position of the hanging block as $y=0$, so the gravitational potential energy immediately after the collision is zero. In terms of the speed $v=v_0/2$ of the blocks after the collision and the mass $m$ of one block, the total energy immediately after the collision is \begin{align} E_i &= \frac{1}{2}(2m) v^2 + \frac{1}{2}kd^2\\ &= \frac{1}{4}mv_0^2 + \frac{1}{2}\frac{m^2g^2}{k}. \end{align} Let $h$ be the distance of the blocks above the equilibrium position of the spring when the blocks are at their maximum height. At this point, the blocks are at rest, so their total energy is \begin{align} E_f &= \frac{1}{2}kh^2 + 2mg(h + d)\\ &= \frac{1}{2}kh^2 + 2mgh + \frac{2m^2g^2}{k}. \end{align} Using conservation of energy, \begin{align} &\frac{1}{4}mv_0^2 + \frac{1}{2}\frac{m^2g^2}{k} = \frac{1}{2}kh^2 + mgh + \frac{2m^2g^2}{k}\\ \rightarrow &\frac{1}{2}kh^2 + 2mgh + \frac{3}{2}\frac{m^2g^2}{k} - \frac{1}{4}mv_0^2 = 0. \end{align} We can solve this quadratic equation for $h$ to obtain \begin{align} h = -\frac{2mg}{k} + \sqrt{\frac{m^2g^2}{k^2} + \frac{mv_0^2}{2k}}. \end{align} In terms of the given numbers, $v_0 = 5.0\,\text{m}/\text{s}$, $m=5.0\,\text{kg}$, and $k=250\,\text{N}/\text{m}$, we get $\boxed{h=15\,\text{cm}.}$

Note that if we set $g=0$ so that there is no gravity, we get \begin{align} h = \sqrt{\frac{mv_0^2}{2k}} = \boxed{50\,\text{cm}.} \end{align} We are left to conclude that the author of the solution was likely in free-fall at the time of its writing.

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  • $\begingroup$ h=15 cm? I got 0.1447. Is this the same thing? Also, how does your equation differ from the one my teacher and I made, which outputs 24 centimeters? $\endgroup$ Dec 2, 2021 at 17:23
  • $\begingroup$ @SebastianPojman-Malo When I plugged the numbers into Mathematica, it spit out 14.5 something, and I rounded to two sig figs. In my answer, I'm taking into account the fact that the spring does not start at its equilibrium length. $\endgroup$
    – d_b
    Dec 2, 2021 at 17:38
  • $\begingroup$ That's interesting. College Board (and I) assume that it absolutely does hit the original block when the spring is at its equilibrium. $\endgroup$ Dec 2, 2021 at 17:45
  • $\begingroup$ No, that is not correct. The hanging block is in equilibrium, which means the spring must be exerting an upward force on the block to keep it from falling under the influence of gravity. So the spring cannot be at its equilibrium length. $\endgroup$
    – d_b
    Dec 2, 2021 at 18:54
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    $\begingroup$ I agree with your h = .145 m. Note that that (h) is measured up from the un-stretched position of the spring and not from the point where the collision occurs. $\endgroup$
    – R.W. Bird
    Dec 6, 2021 at 17:26
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There is already a very good and complete answer from d_b, let me just add a comment to elucidate the possible (erroneous in this case) reasoning by which one could try to forget about gravity.

First, consider one block hanging from the spring in equilibrium. Assume that we provide it with the vertical speed $v$ without any other block sticking to it, so it moves up alone. What height above the initial position will it reach in such a case? Writing energy conservation (for detailed explanation look at d_b's answer, it is quite the same here - only I choose to measure height relative to the initial position of the block for reasons which will become clear shortly), we get:

$$\frac{m v^2}{2} + \frac{m^2 g^2}{2 k} = m g h + \frac{k (h-\frac{mg}{k})^2}{2}$$

Let us play a little with this equation. Open the square in the RHS and observe that $mgh$ cancels:

$$m g h + \frac{k (h-\frac{mg}{k})^2}{2} = mgh + \frac{k(h^2 + \frac{m^2g^2}{k^2} - \frac{2mgh}{k})}{2} = \frac{kh^2}{2} + \frac{m^2 g^2}{2k}$$

Plugging it back to the energy conservation, we observe that $\frac{m^2 g^2}{2k}$ cancels as well, and we are left with

$$\frac{m v^2}{2} = \frac{kh^2}{2}$$

In other words, it looks exactly like we could forget about gravity and initial elongation of the spring altogether: the answer is the same as it would have been for a horizontal spring. This is not a coincidence: what we have derived means that the total (gravitational+elastic) potential energy of this system is quadratic in deviation from the equilibrium, so it really looks like for a horizontal spring. One could also understand it graphically: potential energy of the spring is a parabola with the equilibrium position at the lowest point. If we also add gravity, a linear function is added to this parabola. The result is another parabola with the same shape but another minimum position - at the equilibrium point of the hanging block (equilibrium is always the minimum of total potential energy). If we measure all elongations compared to the new equilibrium, we could forget about both gravity and initial elongation: they compensate each other.

I think this might have been the logic of authors of the solution. However, this logic only works if we measure elongations compared to the position with minimal potential energy. If another block sticks to the first one, the equilibrium position is now shifted below, so the blocks start moving already from a non-equilibrium position and one needs to account for the potential energy of initial position as well - which was not done in the provided College Board solution. As an exercise I would suggest an interested reader to reproduce d_b's answer with this method of total potential taking the above consideration into account.

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  • $\begingroup$ "let me just add a comment to elucidate the possible (erroneous in this case) reasoning by which one could try to forget about gravity." Just for clarity, are you saying that College Board's ignoring gravity is erroneous? Or are you agreeing with College Board? $\endgroup$ Dec 2, 2021 at 17:31
  • $\begingroup$ I disagree with the College Board answer and solution. In my answer I describe a method which they presumably used when solving the problem, but I point out that they forgot to account for the fact that equilibrium position of two blocks on a spring differs from the equilibrium position of one block. $\endgroup$
    – Viking
    Dec 2, 2021 at 17:56
  • $\begingroup$ Walk me through this. (Remember, I'm still in high school...) Does that mean that once the elastic collision happens and the identical block "sticks," the equilibrium position changes in that moment? $\endgroup$ Dec 2, 2021 at 18:00
  • $\begingroup$ Firstly, if they stick, it is called an inelastic collision. Changing equilibrium position means that if one block of mass $m$ stretches the spring in equilibrium by $d$, two such blocks would stretch it by $2d$. So if in this problem we attached the second block to the first one without any velocity, they would move down to reach the new equilibrium position. To prevent confusion, let me stress that the word equilibrium here is used to describe a situation when all forces are balanced and nothing moves (not to be confused with the equilibrium length of the spring - when there are no forces). $\endgroup$
    – Viking
    Dec 2, 2021 at 19:06
  • $\begingroup$ How does conservation of momentum play into this? As I see it, (5kg)(5m/s)=(5+5kg)(v); v=2.5m/s. So according to the conservation of momentum, the two blocks stuck together will move at 2.5m/s immediately following the collision. That being said, they will decelerate because the force of gravity (2mg) will outpower the tension of the spring (mg) at that moment. Because of this, the equilibrium is disturbed, and the amount of compression is reduced. CB's answer assumes equilibrium is NOT disturbed and thus the blocks' velocity does not decelerate, right? $\endgroup$ Dec 2, 2021 at 20:27
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With a 5 kg mass hanging at rest from the spring of constant k = 250 N/m, the stretch of the spring will be X = mg/k =5(9.8)/250 = 0.196 m. With (x) measured positive down from the un-stretched position, and gravitational potential energy chosen to be zero when x = 0, then the energy of the 10 kg mass is (1/2)(10)$2.5^2$ + (1/2)(250)$0.196^2$ - 10(9.8)(0.196) = (1/2)(250)$x^2$ -10(9.8)x. (x when v = 0) Or: 125$x^2$ - 98x – (31.25 +4.802 – 19.208) = 0. Solving gives x = - 0.145 m. (The other solution is for initial velocity downward). Since positive x was measured down, this negative x represents a compression of the spring above the unstretched position. Then the total rise is 0.196 + 0.145 = 0.341 m. For the record, if the moving mass is the same as the hanging mass: then mg = kX and (1/2)m$v^2$ + (1/2)k$X^2$ - (kX)X = (1/2)k$x^2$ - (kX)x. Or rearranged: (1/2)m$v^2$ = (1/2)k$(X^2 – 2Xx + x^2) = (1/2)k(X-x)^2$. Lets try ignoring gravity with x measured from a 10 kg hanging position (which is an additional 0.196m further down): (1/2)(10)$2.5^2$ + (1/2)(250)$0.196^2$ = (1/2)(250)$x^2$. Rearranging: 125$x^2$ = (31.25 + 4.802) Giving x = 0.537. With the starting position 0.196 m above the new equilibrium, this gives a rise of 0.341 m. The bottom line: A mass hanging from a spring defines a new equilibrium position. The kx measured from that position includes the force of gravity (but don't change the mass). Here is an alternative approach: With the stretch of the spring (x) measured positive down from the unstretched position, the net force on the hanging mass is: F = mg - kx Then dF = - k(dx). Integrating both sides gives F(x) - $F_o$ = -k(x - $x_o$). If starting from the new equilbrium then $F_o$ =0.

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  • $\begingroup$ Please consider using MathJax; in its current form this answer is mostly illegible. $\endgroup$
    – rob
    Jan 23, 2022 at 13:55

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