Elastic collision: $v_2 = \frac{2m}{3m}v = \frac{2v}{3}$
Height: $h = v_2^2/2g = (4v^2/9)/2g = 2v^2/9g$
$v^2 = 2g\times45/4$: $h = 2\times2g\times45/4\div9g = 45/9 = \mathbf{5}$ m
Elastic: both momentum and KE conserved ($e=1$). Inelastic: momentum conserved, KE lost ($0 < e < 1$). Perfectly inelastic: objects stick together ($e=0$), maximum KE loss. Super-elastic: $e > 1$ (explosive separation, KE gained from chemical/spring energy). In all collisions: momentum conserved (Newton 3rd law). Only elastic: KE conserved.
Ball 1 ($m_1$, $u_1$) hits ball 2 ($m_2$, $u_2$): $v_1 = \frac{(m_1-m_2)u_1 + 2m_2 u_2}{m_1+m_2}$; $v_2 = \frac{(m_2-m_1)u_2 + 2m_1 u_1}{m_1+m_2}$. Special cases: $m_1 = m_2$: velocities exchange ($v_1 = u_2$, $v_2 = u_1$). $m_1 \ll m_2$ (ball hitting wall): ball bounces back with same speed. $m_1 \gg m_2$: heavy ball continues, light ball moves at $2u_1$.
In CM frame: total momentum = 0. Speeds are reversed after elastic collision. CM velocity: $v_{cm} = (m_1 u_1 + m_2 u_2)/(m_1+m_2)$. Lab frame speeds: $u_1 - v_{cm}$ and $u_2 - v_{cm}$ before; $-(u_1-v_{cm})$ and $-(u_2-v_{cm})$ after (elastic). Back-transform to lab frame. CM frame simplifies elastic collision analysis significantly.
Classic experiment to measure bullet speed: bullet of mass $m$ and speed $v$ embeds in pendulum block of mass $M$. Perfectly inelastic: $mv = (m+M)V$ → $V = mv/(m+M)$. Pendulum swings to height $h$: $(m+M)gh = \frac{1}{2}(m+M)V^2$ → $h = V^2/2g$. So bullet speed $v = (m+M)\sqrt{2gh}/m$. Demonstrates momentum conservation + energy methods.
For a one-dimensional elastic collision between masses $m_1$ and $m_2$, solving momentum and kinetic energy conservation together gives $v_1 = \frac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2}$ and a mirror expression for $v_2$. Three special cases are worth memorising: equal masses simply exchange velocities; a light body striking a very heavy one rebounds with nearly the same speed; and a heavy body striking a light one barely slows while the light one leaves at nearly twice the incoming speed.
Defined as $e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}$, it classifies every collision on one scale: $e = 1$ is perfectly elastic, $e = 0$ is perfectly inelastic, and real collisions fall between. For a ball dropped from height $h$ and rebounding to $h_1$, $e = \sqrt{h_1/h}$, and after $n$ bounces the height is $h e^{2n}$ — a result that turns a bouncing-ball question into one line.
When the collision is oblique, momentum must be conserved separately along two perpendicular axes, giving two equations instead of one. The usual choice is to put one axis along the line of impact and the other perpendicular to it, because for smooth bodies the perpendicular component of each velocity is unchanged. A result worth remembering: when a moving body collides elastically with an identical stationary body in two dimensions, the two always separate at right angles to each other. This is why a cue ball and an object ball in carrom or billiards move apart at 90° after a non-head-on elastic hit.
Applying kinetic energy conservation to an inelastic collision. Momentum is conserved in every collision; kinetic energy only in elastic ones. Check which type before writing the second equation.
Ignoring direction. Momentum is a vector. In a head-on collision one velocity must carry a negative sign, and dropping it is the most common arithmetic error here.