Question bank — Spring-mass systems — collision problems

False as a general claim — the spring force is the contact force, and at the very first instant the spring is barely squished, so that force $kx$ (with $x\approx 0$) and its impulse are tiny; almost no momentum or energy has been exchanged yet. You track the encounter by momentum, and KE is only conserved across the whole encounter because the ideal spring later returns everything it stored.

False — it is maximal when both blocks share the same velocity ($v_{cm}$), i.e. relative velocity is zero. The incoming block is usually still moving then.

True — an ideal spring stores KE then returns 100% of it, so the before-and-after KE match, which is the definition of elastic (see Elastic and Inelastic Collisions).

False — the wall supplies an external horizontal force, so momentum is not conserved; only mechanical energy is.

False — the whole system still drifts at $v_{cm}$, so it keeps the drift KE $\frac{1}{2}(m_1+m_2)v_{cm}^2$. Only the relative-motion KE, $\frac{1}{2}\mu (u_1-u_2{)}^2$, is stored in the spring.

True — the blocks are locked together moving at $v_{cm}$, and the stored $\frac{1}{2}\mu (u_1-u_2{)}^2$ is trapped forever, which is exactly the inelastic signature.

False — $x_{max}=|u_1-u_2|\sqrt{\mu /k}$ depends on the relative speed and the symmetric $\mu$ (swapping $m_1\leftrightarrow m_2$ leaves $\mu$ unchanged), so the depth is the same.

False — with zero relative velocity from the start, the gap never shrinks, so $x_{max}=|u_1-u_2|\sqrt{\mu /k}=0$. Nothing compresses.

The error is timing: you cannot equate initial KE to spring PE at first contact, where $x\approx 0$ and almost nothing has been exchanged. Energy is only clean once you correctly track the stored PE across the full compression and subtract the drift KE.

Wrong — the system still moves at $v_{cm}$, carrying drift KE that can't be stored. Only the CM-frame part $\frac{1}{2}\mu (u_1-u_2{)}^2$ becomes spring PE; you must subtract $\frac{1}{2}(m_1+m_2)v_{cm}^2$.

The target's speed peaks after max compression, when the spring is pushing it forward. Compression peaks at equal velocities, before the spring re-expands.

No — an ideal spring returns exactly the energy it stored, so the block leaves at the same speed it arrived (reversed direction). Gaining speed would violate energy conservation.

There is no second free body to share momentum with, and the wall makes momentum non-conserved, so that equation is meaningless here. The correct tool is energy: $\frac{1}{2}m{u}^2=\frac{1}{2}k{x}^2$.

No energy is lost; $\mu$ is simply the correct effective mass for relative motion. The "missing" KE isn't destroyed — it's the drift KE $\frac{1}{2}(m_1+m_2)v_{cm}^2$ that was never available to store.

A light block striking a heavier one genuinely rebounds; the elastic result $u_1^{\prime }=\frac{m_1-m_2}{m_1+m_2}{u}_1$ is negative when $m_1<m_2$, and the minus sign just means leftward. Momentum and energy both check out, so the rebound is real.

Because different forces dominate at different timescales — the huge, brief contact force needs momentum (impulse view), while the slow, finite spring force needs energy. One law per phase, chosen by which force is negligible.

$v_{cm}$ is about the whole system's drift, so it uses total mass; $x_{max}$ is about the blocks' relative squeezing, and relative motion behaves like a single body of mass $\mu$. See Reduced Mass and Two-Body Problems.

Because "ideal spring returns all stored energy" is the physical content of restitution $=1$; the spring mechanism guarantees the elastic result by construction.

At that instant both blocks share one velocity, turning $m_1{u}_1+m_2{u}_2=(m_1+m_2)v$ into a single-unknown equation — the cleanest snapshot the problem offers.

Inelastic collisions convert KE into heat/deformation (unrecoverable); an ideal spring only borrows KE as recoverable elastic PE and hands it all back.

In that frame the total momentum is zero, so at max compression everything is momentarily still and all the frame's KE — exactly $\frac{1}{2}\mu (u_1-u_2{)}^2$ — sits in the spring.

The compression phase up to the turning point is identical physics either way; latching or releasing only decides what happens after the peak, not how deep the squeeze got.

Zero — the relative approach speed $|u_1-u_2|=0$, so $x_{max}=|u_1-u_2|\sqrt{\mu /k}=0$; they simply coast together.

$x_{max}=|u_1-u_2|\sqrt{\mu /k}\to 0$ — an infinitely stiff spring barely compresses, and the encounter looks like an instantaneous elastic bounce.

$\mu \to m_1$ and $v_{cm}\to 0$, recovering the wall case: $x_{max}=|u_1|\sqrt{m_1/k}$ with all of block 1's KE stored.

The blocks swap velocities — the incoming block stops dead and the target moves off at $u_1$, since $u_1^{\prime }=\frac{m_1-m_2}{m_1+m_2}{u}_1=0$.

Since block 1 is falling behind, its face never presses the free end of the spring, so no compression occurs and the blocks simply drift apart unchanged. (If instead the spring were glued to both blocks, moving apart would stretch it — a different, tension problem outside this note.)

Mathematically $x_{max}\to \infty$, but that's an idealisation: a real spring has a finite length and coils that eventually touch (or the blocks collide), so the formula only holds while the spring behaves ideally. Practically, a very soft spring means a huge but bounded squeeze and a very gentle, slow interaction.

Zero — total momentum cancels, so at max compression both are momentarily at rest and the entire initial KE, $\frac{1}{2}\mu (2u_1{)}^2$, is stored (the Centre of Mass Frame and lab frame coincide here).