Following the lecture “2D collision experiment”. Do each Try it to see it on screen.
1. Momentum is conserved component by component
During the brief collision the two balls push on each other with equal and opposite forces, and other horizontal forces are negligible. So the x and y components of momentum in the horizontal plane are each conserved. The target starts at rest; the incident ball arrives along x.
Try it Roll experiment 1 once and experiment 2 once and read “sum after” and “difference” in the table. Without scatter both the p_x difference and the p_y sum are close to 0.
2. Equal height, horizontal launch: equal fall time
Both balls leave horizontally from height h. Vertically they fall freely from rest; horizontally they move uniformly. The fall time depends only on h, so the distance r on the floor gives the speed without a speedometer.
t=\sqrt{\frac{2h}{g}},\qquad v=\frac{r}{t}=r\sqrt{\frac{g}{2h}}
Try it Step through the side view: both balls are at the same height at the same moment. With h=80\,\mathrm{cm}, t\approx0.404\,\mathrm{s}. A different stand height changes the distances but not v=\frac{r}{t}.
3. Experiment 1: speed and energy of a rolling ball
Rolling the incident ball alone gives v_1 from r_1. A ball rolling without slipping turns part of its potential energy into rotation, so its translational kinetic energy is less than m_1gH.
m_1gH=\frac{1}{2}m_1v_1^2+\frac{1}{5}m_1v_1^2\quad\Rightarrow\quad E_1=\frac{1}{2}m_1v_1^2=\frac{5}{7}m_1gH
Try it With H=30\,\mathrm{cm}, experiment 1 gives r_1\approx82.8\,\mathrm{cm} and v_1\approx2.05\,\mathrm{m/s}; the table note says E_1 is about 71\% of m_1gH. Switch to sliding: r_1\approx98.0\,\mathrm{cm} and 100\%.
4. The sideways offset sets the angle
At contact the centres are 2R apart, and the force between smooth balls acts along the line of centres. The target leaves along that line, at an angle set by its sideways offset b.
\sin\theta_2=\frac{b}{2R}
b=0 is a head-on (one-dimensional) collision, b=R gives \theta_2=30^\circ, and b near 2R only grazes the target.
Try it Watching the inset, set b to 0, 6.3\,\mathrm{mm}, 12.5\,\mathrm{mm}, 25\,\mathrm{mm} (R=12.5\,\mathrm{mm}): \theta_2 reads 0^\circ, 14.6^\circ, 30^\circ, 90^\circ.
5. Equal masses, elastic: \theta_1+\theta_2=90^\circ
With m_1=m_2, momentum gives \vec{v}_1=\vec{v}_1'+\vec{v}_2' and an elastic collision gives v_1^2=v_1'^2+v_2'^2. The vector triangle obeys Pythagoras, so \vec{v}_1' and \vec{v}_2' are perpendicular. Head-on, the incident ball stops and the target leaves with v_1.
\vec{v}_1=\vec{v}_1'+\vec{v}_2',\qquad v_1^2=v_1'^2+v_2'^2\quad\Rightarrow\quad\theta_1+\theta_2=90^\circ
Try it With the defaults (b=R) experiment 2 gives \theta_1=60^\circ, \theta_2=30^\circ, r_1'\approx41.4\,\mathrm{cm}, r_2'\approx71.7\,\mathrm{cm}. With b=0, P_1' lands on O_1.
6. Changing masses and restitution
A heavier target deflects the incident ball more; head-on with m_2>m_1 the incident ball bounces back toward the ramp (hard to record in a real lab). With restitution e<1 kinetic energy is lost and, even for equal masses, \theta_1+\theta_2 drops below 90^\circ. Momentum is conserved in every case.
v_{2}'=\frac{(1+e)\,m_1}{m_1+m_2}\,v_1\cos\theta_2
Try it At e=0.80, \theta_1+\theta_2\approx80^\circ and the energy difference in the table turns negative. With m_2=40\,\mathrm{g} and b=0 a message says the incident ball bounced back.
7. Checking conservation with distance vectors
All balls share the fall time t, so in m\vec{v}=\frac{m\vec{r}}{t} the t cancels. Join m\vec{r} arrows on the paper: a closed triangle means momentum is conserved. For an elastic collision, squared distances check energy too.
Try it In the vector panel, check that the grey m_1\vec{r}_1 ends where the orange m_1\vec{r}_1' plus the teal m_2\vec{r}_2' ends. With scatter on and several runs the ends separate slightly.
8. Two origins for measuring
The plumb line marks the target's start O_2. At contact the incident ball's centre is at O_1, 2R from O_2 on the side opposite to the target's departure. So r_1, r_1', \theta_1 are measured from O_1 and r_2', \theta_2 from O_2; using one origin for both distorts r_2' and \theta_2.
Try it Increase R and confirm in the inset that O_1 and O_2 are 2R apart; the marks O_1, O_2 on the paper move too.
9. Repeats and error: “conserved to what extent?”
In a real lab the release height, direction, stand position and reading vary a little each time, so marks scatter. Roll at least three times, use the centre of each group, and treat the spread as the measurement uncertainty. Conclude “conserved within this error”, not just “conserved / not conserved”. Real balls carry rotational energy and feel friction on the ramp and stand, so energy conservation is checked less precisely than momentum.
Try it Turn on experimental scatter and use Roll 3× for experiments 1 and 2. Compare the spread of the marks (about 1–2\,\mathrm{cm}) with the percentage differences in the table.