Spring-mass systems — collision problems: Foundations

This is the toolbox page for Spring–Mass Collision Problems. We assemble the tools in the order they get used.

1. Mass $m$ — "how hard to shove"

The picture: think of two boxes, a small one and a fat one. Push each with the same hand-force — the fat one speeds up slower. That reluctance to speed up is mass.

Why the topic needs it: collisions are all about how two masses share motion. Big mass barely changes; small mass gets flung. The letters $m_{1}$ (the mass of block 1) and $m_{2}$ (block 2) are just name-tags on the two boxes.

2. Velocity $u$ and $v$ — "speed WITH a direction"

We must pick a direction to call positive before we write any number. The parent note says "take rightward positive." That single choice is what lets a velocity be negative — a negative velocity just means "moving left".

Figure — Spring-mass systems — collision problems

3. Momentum $p = m v$ — "quantity of motion"

The picture: imagine each block dragging a bar behind it whose length is $m$ and whose direction is the velocity. Momentum is that bar. When two blocks collide, the total length of all bars, added with their signs, does not change — that's the law we lean on.

Why $m_{1} u_{1} + m_{2} u_{2}$ and not $m_{1} u_{1} - m_{2} u_{2}$ ? You never subtract on purpose — the sign lives inside each velocity. If block 2 moves left, $u_{2}$ is already negative, so $m_{2} u_{2}$ is already negative. Always add; let the signs do the work.

4. Impulse — "why the spring is invisible during the hit"

The picture: a change in momentum is like filling a bucket. A firehose (huge force, short time) and a dripping tap (tiny force, long time) can fill the same bucket. During a collision the contact force is a firehose — enormous but over a blink of time.

Why this matters here: the spring force $k x$ is a dripping tap during that blink — finite force, near-zero time → near-zero impulse → it barely changes momentum. That is the whole reason we're allowed to use momentum (not energy) during the collision. Get this and the "MEME" mnemonic makes sense: Momentum during the Moment.

5. Kinetic energy $\frac{1}{2} m v^{2}$ — "the motion-fund"

Why the $\frac{1}{2}$ and the square? Doubling the speed quadruples the KE (that's the square); the $\frac{1}{2}$ is the exact bookkeeping factor that makes energy balance with force-times-distance. You don't need to derive it here — just trust that $\frac{1}{2} m v^{2}$ is the "size of the motion-fund".

6. Spring stiffness $k$ and elastic PE $\frac{1}{2} k x^{2}$ — "the energy bank"

The spring pushes back with force $k x$ — the more you squish, the harder it shoves. That force is finite (never a firehose), which is exactly why it's ignorable during the collision instant.

The picture: the spring is a piggy bank. Kinetic energy goes in as it squishes ( $x$ grows), then comes back out as it un-squishes. An ideal spring is a piggy bank with no hole in the bottom — it returns every joule.

7. The common velocity $v_{c m}$ — "the moment they match speed"

Because momentum is conserved from just-after-contact to that instant: $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v \Rightarrow v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$

Read the formula in words: total momentum divided by total mass = the velocity of the whole system's balance-point. This is why it's called the centre-of-mass velocity — see Centre of Mass Frame.

8. Reduced mass $μ$ — "the effective mass of the GAP"

Feel for the number: $μ$ is always smaller than either mass. If one block is a wall (infinite mass), $μ \to$ the other block's mass. If both are equal $m$ , then $μ = m /2$ .

Why it earns its place: the energy the spring can store depends only on the approach speed $(u_{1} - u_{2})$ and this effective gap-mass $μ$ : $stored in spring \frac{1}{2} k x_{ma x}^{2} = relative-motion KE \frac{1}{2} μ (u_{1} - u_{2})^{2} \Rightarrow x_{ma x} = (u_{1} - u_{2}) \frac{μ}{k}$

The chunk $\frac{1}{2} (m_{1} + m_{2}) v^{2}$ (the whole system drifting at $v$ ) can never be stored — the system is still moving as one lump. Only the approaching part converts. That is the single most-missed idea in the parent note.

I can write the momentum of one block and of two blocks

p = m v

; total

= m_{1} u_{1} + m_{2} u_{2}

, always added with signs baked in.

I can explain why the spring is "invisible" during the collision instant

Its force is finite but the time is near-zero, so its impulse (force × time) ≈ 0 and momentum is unchanged.

I can write kinetic energy and say why it's never negative

K E = \frac{1}{2} m v^{2}

; the velocity is squared so direction cancels.

I can write elastic PE and describe the spring as a bank

P E = \frac{1}{2} k x^{2}

; an ideal spring stores every joule and returns it all.

I can state when compression is maximum

When both blocks share one velocity

v

(relative velocity

= 0

I can compute the common velocity

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}

— total momentum over total mass.

I can compute reduced mass and say why it's used

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

; it's the effective mass of the relative motion the spring actually stores.

I can extract

x_{ma x}

using a square root

From

\frac{1}{2} k x_{ma x}^{2} = \frac{1}{2} μ (u_{1} - u_{2})^{2}

, take the root:

x_{ma x} = (u_{1} - u_{2}) μ / k

Foundations — Spring-mass systems — collision problems

1. Mass $m$ — "how hard to shove"

2. Velocity $u$ and $v$ — "speed WITH a direction"

3. Momentum $p = m v$ — "quantity of motion"

4. Impulse — "why the spring is invisible during the hit"

5. Kinetic energy $\frac{1}{2} m v^{2}$ — "the motion-fund"

6. Spring stiffness $k$ and elastic PE $\frac{1}{2} k x^{2}$ — "the energy bank"

7. The common velocity $v_{c m}$ — "the moment they match speed"

8. Reduced mass $μ$ — "the effective mass of the GAP"

9. The square-root — "undo the square"

How the tools feed the topic

Equipment checklist

Connections