# How/ Why is spring compression maximum when the two blocks have equal velocities?

Consider a block of mass $m_1$ attached to a spring. It's advancing with a slower velocity $v_1$ as compared to that of an approaching block of mass $m_2$'s velocity $v_2$ on the friction-less surface. Obviously, the two blocks will collide.

Now, I have read at several places that the spring compression will be maximum when the two blocks have equal velocities. But I couldn't understand that. It's somewhat counter-intuitive to me because first block is accelerating and the one behind is decelerating so how does that eventually lead to maximum compression in the spring?

Research effort: I asked a friend on the chat room but his explanation was unsatisfactory. I searched for youtube experiments on this (they help immensely in developing concepts intuitively) but couldn't find any. My books too explain this theoretically without any diagrams.

Please supplement your answer with diagrams/ mathematical proof to explain this concept (and other related concepts like maximum elongation ,if possible).

• The simple thing to see is that from the reference of either block, both blocks need to have a velocity of 0 with respect to each other when the spring is compressed the most, this directly translate to both having the same velocity in reference to another object. Sep 8, 2017 at 22:14
• Compressing the spring requires velocity with respect to each other. Sep 8, 2017 at 22:48
• If you just have two blocks moving towards each other, there is no upper limit to how much a spring between them can be compressed (well, assuming that the spring doesn't bottom out). You can get more and more spring compression by increasing the velocity of one block or the other or both towards each other without limit. I think that the question is supposed to have some constraint like "assume that the sum of the speeds of both blocks is constrained to be a constant". Then (for blocks of equal masses) the spring compassion will be maximum if the speeds are equal and opposite.
– user93237
Sep 8, 2017 at 23:01
• Where did you read this? What was unsatisfactory about the explanation you were given? Sep 9, 2017 at 8:07

Let $m_1,m_2$ have velocities $v_1,v_2$, where $v_1<v_2$ and $m_1$ is initially ahead of $m_2$.

$m_2$ has a spring attached to its front (in direction of motion).

Once $m_2$ catches up with $m_1$, the spring will be compressed between the two masses, and will (in an effort to return to its initial state) exert a force on both $m_1$ and $m_2$.

This force will cause $m_1$ to accelerate, and $m_2$ to decelerate.

Prior to the closest approach of the two masses, $v_1<v_2$. We know this because (given that the masses are going to get closer), $m_2$ must still be approaching $m_1$.

Following the closest approach of the two masses, $v_1>v_2$. We know this because, given that the masses are no longer getting closer together, and they are still accelerating and decelerating, respectively, $m_2$ must be receding from $m_1$.

Given that we know what is happening on either side of the closest approach of the masses, and we know that the velocities of the masses must change without instantaneous shift (cannot skip past the point at which their velocities are equal), the only logical conclusion is that the velocities of the two masses are equal at the closest approach.

• But $m_2$ is decelerating and $m_1$ accelerating. Doesn't that effect the separation (i.e. increase it)? Sep 9, 2017 at 4:29
• @Abcd relative acceleration of the two masses does not directly affect the separation, it affects the rate of separation. Initially the rate of separation is negative, as the masses become closer together, and then, as the spring causes them to accelerate, the rate of separation increases from negative towards positive. You can see from this that there will be a point at which the rate of separation reaches 0, and this point will be the closest approach. samjoe described the situation using rate of separation in his posted answer. Sep 9, 2017 at 4:45
• How is rate of separation negative initially? Sep 9, 2017 at 5:13
• The rate of separation would be defined as change in separation / time. Initially the change in separation would be negative, as the distance between the two decreased (assuming that you would defined increased separation as increased distance between the two); therefore, the rate of separation would be negative. Sep 9, 2017 at 5:17

Although it should be obvious, and physical explanations as given in other answers are always much better, you can verify it by doing simple calculations. Let:

• displacement of $m_1$ be $\vec{x_1}$
• displacement of $m_2$ be $\vec{x_2}$

Now, the compression/elongation in spring will be $X=|\vec{x_2} - \vec{x_1}|$.
Extrema occurs when $\frac{dX}{dt} =0$

$$\frac{dX}{dt} =\dfrac{d}{dt}\vec{x_2}-\dfrac{d}{dt}\vec{x_1} =0$$

or $\vec{v_1} = \vec{v_2}$

As explained in other answers, this value of extension $X$ is achievable, as the spring is exerting equal force on each body but in opposite direction (always)

• How did you differentiate inside an absolute value sign? You completely ignored the modulus in your second step Sep 11, 2017 at 23:03
• @xasthor sorry for late reply! We are dealing with 1D motion and I have taken extension and compression both as positive. Whichever way you resolve the modulus, result will be same. But I too think taking $X$ as $\vec{x_2} - \vec{x_1}$ and treating compression and extension as of being different signs is more proper. Sep 15, 2017 at 18:22

When the spring can't be compressed any more, they have to stop relative to each other.

Take the view of the center of mass frame, in which only differential velocities remain. Only changes in the distance between the blocks affect the spring (and vice versa).

Imagine that the spring is attached to observer $A$.
An object is thrown at observer $A$.
The object hits observer $A$.

When will observer $A$ observe that the compression of the spring is greatest?
It will be when the object is not moving relative to observer $A$.
(If the object is moving towards observer $A$ the spring is being compressed more and if the object is moving away from observer $A$ the spring is being extended more.)

This means that any other observer who observes the object hitting observer $A$ will note that maximum compression occurs when the object and observer $A$ are moving at the same velocity relative to observer $B$.