# Homework Help: Two blocks and a spring

1. May 6, 2010

### Meithan

I'm a TA in an introductory course of (classical) mechanics at the university. I've been baffled for the couple two hours at this problem, taken from Resnick, Halliday & Krane (Fourth Edition in english; it's the 3rd ed. in spanish). I initially solved the problem, thinking my reasoning was sound, and was surprised when I discovered my (numerical) answer was not the same as the book's.

After more than two hours of checking the reasoning, the algebra, and numerical evaluations, I still find a numerical solution distinct from the one in the book. So I'm wondering whether the book's got it wrong, or whether I'm missing something in the solution. Anyway, here's the problem and my solution.

Problem (translated from spanish)

A block of mass $$m_1=1.88$$ kg slides along a frictionless table at a speed of $$10.3$$ m/s. In front of it, and moving in the same direction, is a block of mass $$m_2=4.92$$ kg, moving at a speed of $$3.27$$ m/s. A massless spring with a spring constant $$k=11.2$$ N/cm is attached to the second block, facing the first block, as shown in the figure. When the blocks collide, what is the maximum compression of the spring?

http://img525.imageshack.us/img525/8029/fig.png [Broken]

My solution

I assumed that when the spring is at maximum compression, the two blocks have momentarily the same speed, as if it were a completely inelastic collision. Alternatively, one can also hop in the center-of-mass frame, and it is clear (if we consider them point masses) that at this moment both blocks are at rest (in this frame), thus having the same velocity. (And if two objects have the same velocity in a given inertial reference frame, they'll have the same velocity in any inertial reference frame - even in SR.)

Then, we can write the conservation of total linear momentum as

$$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_3$$

where $$v_1$$ and $$v_2$$ are the initial velocities of the blocks, respectively, and $$v_3$$ is their common velocity at the exact instant the spring is maximally compressed. We can directly solve this for $$v_3$$:

$$v_3 = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$

Now, even if the collision is not elastic, the total energy of the system (blocks+spring) must be conserved. The kinetic energy is not conserved because part of it is being transfered to the spring as potential energy. We know the speeds of the blocks initially and at maximum compression, so we can in principle solve the problem for the maximum compression. The conservation of total energy can be stated in this case as:

$$\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (m_1+m_2) v_3^2 + \frac{1}{2} k x^2$$

where $$x$$ is the maximum compression of the spring. We know all quantities in this equation except for $$x$$, which is what is asked (we've calculated $$v_3$$ in terms of known quantities). Therefore, we can solve for $$x$$. After some algebra (which I have thoroughly checked, both by doing it myself several times and also by doing it with Maxima), I get:

$$x = \left|v_1 - v_2\right| \sqrt{\frac{m_1 m_2}{k(m_1 + m_2)}}$$

A sign ambiguity in the algebraic derivation is produced when we take the square root, so I've taken the absolute value of $$v_1-v_2$$ because we expect a positive physical answer.

Evaluating the solution with the numbers given in the problem,

$$m_1=1.88$$ kg
$$m_2=4.92$$ kg
$$v_1=10.3$$ m/s
$$v_2=3.27$$ m/s
$$k=1120$$ N/m (converted from the given 11.2 N/cm)

I get (and I've also done the numerical computation many times on the computer):

$$x \approx 0.245$$ m, or some 24.5 cm.

However, the solution in the book is 35.9 cm! The given answer has enough significant figures to make it unlikely that it was a printing error.

What is even more baffling is that a friend of mine found the same problem in a newer edition of the book, only with different numbers, but my solution still gives the wrong answer in that edition!

Is there an error in my reasoning? Could the book's numerical answer be wrong in both editions?

Last edited by a moderator: May 4, 2017
2. May 6, 2010

### rl.bhat

The conservation of total energy can be stated in this case as
1/2*m1*vi^2 + 1/2*m2*vi^2 = 1/2(m1+m2)v3^2 + 1/2k*x^2 + 1/2*m2*(v3-v2)^2- 1/2*m1*(v1-v3)^2
where the last two term are the fall of KE of m1 and gain in KE of m2 before they reach the final velocity v3.

3. May 7, 2010

### Meithan

I don't see why you'd introduce such an arbitrary modification of the equation of energy. To me it's simple: (total mechanical energy at instant 1) = (total mechanical energy at instant 2), if there are no losses. The total mechanical energy before the collision is in the form of the kinetic energy of the blocks, moving at their initial speeds $$v_1$$ and $$v_2$$. At maximum compression, the total energy is in the form of potential energy stored in the spring plus the kinetic energy of the blocks at that moment (they've changed their speeds, but they aren't at rest - they can't be because of conservation of linear momentum).

One way to state this in this case is to say that the difference in the total kinetic energy of the blocks before the collision and at max. compression is equal to the energy stored in the spring (if one only looks at the kinetic energy balance, this is what makes the collision inelastic, in a way):

$$-\Delta K = \frac{1}{2}k x^2$$

where, of course,

$$\Delta K = K_f - K_i = \frac{1}{2}m_1 v_f^2 + \frac{1}{2}m_2 v_f^2 - \left( \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \right)$$

which is algebraically equivalent to what I've written (the minus sign is because the energy stored in the spring is the loss of kinetic energy of the blocks).

Now, while I disagree with your suggestion on a physical basis, I will try to compute the final answer taking it into account, to see if that matches the book.

Edit:

I've run the numbers through your formula. I didn't obtain the algebraic solution for it's too much work. I simply calculated $$v_3$$ using the expression in my original post, which gives $$v_3=5.21$$ m/s, and then used that in your statement of the conservation of energy. I get x=29.5 cm, which is closer to the answer but it's still not it.

Last edited: May 7, 2010
4. May 13, 2010

### Meithan

I finally solved my issue: the answers in the books are wrong! I found the same problem (only with different numbers) in the latest edition of the book (8th), and this time my analytical solution gives the correct answer. Furthermore, the solution of the problem is available online as an http://higheredbcs.wiley.com/legacy/college/halliday/0471758019/ilw/c9_p53.htm" [Broken]. It follows the same reasoning as I did and arrives at the same two equations. Here's a screenshot of the final state:

[PLAIN]http://img80.imageshack.us/img80/269/probsol.png [Broken]

(And yes, disappointingly, the energy equation in the interactive solution has a small error: the first term should contain $v_{1i}$ instead of $r_{1i}$. It's only a typing error as it's clear that they intended to mean $v_{1i}$)

Last edited by a moderator: May 4, 2017
5. May 13, 2010

Thank you.