Question bank — Elastic collisions — 2D - angle relationship

False — it needs equal masses with the target at rest. Unequal masses leave a nonzero ${\vec{v}}_1\cdot {\vec{v}}_2$, so the opening angle drifts off $90^{\circ }$.

True — momentum conservation only needs zero external force during impact. It says nothing about energy, so it survives inelastic hits.

False — the proof subtracts the energy equation, which inelastic collisions violate; the opening angle comes out less than $90^{\circ }$.

False on the side: its angle has magnitude $60^{\circ }$ but points on the opposite ($-y$) side, i.e. its signed direction is $-60^{\circ }$. Their y-momenta ($+v_1\sin {\theta }_1$ and $-v_2\sin {\theta }_2$) then cancel back to the original zero.

True — that is exactly the kinetic-energy equation after masses cancel; it is the Pythagoras statement of the right triangle. (See Kinetic Energy for why energy is a single scalar, not a vector.)

False — the cue ball stops, so there is no second direction; ${\vec{v}}_1\cdot {\vec{v}}_2=0$ is trivially true but the opening angle is undefined (see the head-on definition above).

False — heavy hits light gives opening angle less than $90^{\circ }$; the heavy ball barely deflects. Light-hits-heavy is the one that exceeds $90^{\circ }$.

False — $u$ cancels completely. The perpendicularity is purely geometric; any speed gives the same right-angle split.

Error: momentum in 2D has only two components (x and y) → 2 equations. With energy that is 3 equations total, not 4.

Error: you may not add magnitudes of vectors. Only when they are parallel does that hold; here they meet at an angle, so $u_1^2=v_1^2+2{\vec{v}}_1\cdot {\vec{v}}_2+v_2^2$.

Error: squaring gives the cross term $2{\vec{v}}_1\cdot {\vec{v}}_2$. It vanishes only after subtracting the energy equation — the cancellation is the whole proof, not an assumption. (See Dot Product for why $\vec{a}\cdot \vec{b}=0$ means perpendicular.)

Error: it leaves on the opposite ($-y$) side, so its y-momentum is $-m{v}_2\sin {\theta }_2$. With our y-axis fixed, opposite sides mean opposite signs — and only then does total y-momentum stay zero.

Error: KE has no direction — it is one scalar number. It contributes exactly one equation, never split into components.

Error: backscatter of the incoming ball happens when it hits a heavier target (light-hits-heavy), pushing the opening angle toward but not reaching a full split; a heavy incomer never backscatters. (See Elastic collisions — 1D for the mass-ratio rules behind bounce-back.)

Squaring (dotting with itself) turns the vector subtraction into magnitudes plus a single ${\vec{v}}_1\cdot {\vec{v}}_2$ term — precisely the quantity whose vanishing means perpendicular.

Both the momentum equation ($m\vec{u}=m{\vec{v}}_1+m{\vec{v}}_2$) and the energy equation ($\frac{1}{2}m{u}^2=\dots$) carry a common $m$ on every term that divides out, leaving a purely kinematic relation.

We fixed the y-axis once. The cue ball scatters to $+y$, the target to $-y$; a velocity pointing along $-y$ has a negative y-component by definition, so its contribution is $-v_2\sin {\theta }_2$. Both signs must be opposite to keep the total at the initial zero.

An angle is defined between two directions. If one velocity is zero (head-on stop), there is no second direction, so "$90^{\circ }$ between them" is empty.

Elastic means $\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$, i.e. $u^2=v_1^2+v_2^2$ — the exact form of the hypotenuse-legs relation.

Lost kinetic energy shortens the outgoing legs relative to the incoming hypotenuse; to still add (as vectors) to $\vec{u}$, the legs must lean toward each other, narrowing the V below $90^{\circ }$.

The impact parameter (sideways offset of the hit) only sets how the $90^{\circ }$ is shared (${\theta }_1$ vs ${\theta }_2$); conservation laws fix their sum at $90^{\circ }$ regardless.

Momentum alone allows any split of $\vec{u}$ into two vectors; only adding energy (the length constraint) pins the cross term to zero. (See Conservation of Momentum for why the vector sum is fixed.)

The cue ball barely deflects (${\theta }_1\to 0$), so the struck ball goes nearly perpendicular (${\theta }_2\to 90^{\circ }$) and moves slowly — the V opens to a right angle with one very short leg.

Undefined — the cue ball stops and the target takes all the velocity; there is only one moving direction, so no angle exists (this is the single degenerate case defined above, not a $0^{\circ }$ or $90^{\circ }$ value).

No — that is the extreme light-hits-heavy case; the incoming ball reflects and the target essentially doesn't move, so no equal-mass right angle exists.

In the CM frame equal-mass balls always emerge exactly opposite ($180^{\circ }$); the neat $90^{\circ }$ is a feature of the lab frame where the target starts at rest. (See Center of Mass Frame for how switching frames changes the picture.)

The $90^{\circ }$ rule fails — it was derived assuming zero initial momentum for $m_2$. With both moving you must return to the full component equations.

No — for equal masses, target at rest, and both balls moving, ${\vec{v}}_1\cdot {\vec{v}}_2=0$ is exact; the only exceptions are the degenerate head-on and grazing limits, not an intermediate angle. (Sub-$90^{\circ }$ opening angles live in Inelastic collisions, where energy is lost.)