(a) $\theta=\pi-2\varphi=2\arccos\dfrac{b}{R+r}$, so $b=(R+r)\cos\dfrac{\theta}{2}$. (b) $dP=\tfrac12\sin\theta\,d\theta$. (c) $P(\theta<\tfrac{\pi}{2})=\tfrac12$, whatever $R$ and $r$ are.
(ii) Problem Statement
"A stationary ball of radius $R$ is irradiated by a parallel stream of particles whose radius is $r$. Assuming the collision of a particle and the ball to be elastic, find: (a) the deflection angle $\theta$ of a particle as a function of its aiming parameter $b$; (b) the fraction of particles which after a collision with the ball are scattered into the angular interval between $\theta$ and $\theta+d\theta$; (c) the probability of a particle to be deflected, after a collision with the ball, into the front hemisphere."
The aiming parameter $b$ is the sideways distance between the ball's centre and the particle's centre line. An elastic collision loses no kinetic energy. The "front hemisphere" means the particle still moves somewhat forward, $\theta<\pi/2$.
(iii) The Big Idea in Plain Words
Think of a smooth ball bouncing off a smooth wall. The particle touches the ball when their centres are $R+r$ apart. So we can shrink the particle to a point and grow the ball to radius $a=R+r$.
The bounce is a mirror bounce. The particle's sideways speed stays. Its speed toward the ball's centre flips. That fixes the new direction.
Particles are spread evenly over the beam. A ring of the beam with radius between $b$ and $b+db$ holds a share of them equal to its area. Matching rings to angles gives the angle distribution.
(iii-b) What Is This in Atomic and Nuclear Physics?
It is the idea of a cross-section. A target has an effective area that the beam must hit to be scattered. For a hard ball of touch radius $a=R+r$ that area is $\sigma=\pi a^2$. Counting how many particles are scattered tells us $\sigma$, and so tells us the size of the target. This is how scattering experiments measure the size of nuclei.
The result $dP=\tfrac12\sin\theta\,d\theta$ means isotropic scattering. The angle window $\theta$ to $\theta+d\theta$ covers a ring of solid angle $2\pi\sin\theta\,d\theta$. So the share per unit solid angle is the same in every direction: $d\sigma/d\Omega=a^2/4$. A hard ball scatters equally into the front and back hemispheres, so $P=\tfrac12$. Doubling the radius makes $\sigma$ four times bigger (more particles hit), but it spreads them in the same pattern. This is why the graphs of $dP$ and $P$ do not change when you change $R$ or $r$. Only $b$ against $\theta$ changes, because it depends on $b/a$.
Compare with Coulomb scattering (Irodov 6.2). For charges the push reaches out, so the pattern depends on energy and charge and is strongly peaked in the forward direction, $d\sigma/d\Omega\propto1/\sin^4(\theta/2)$. A flat pattern means a hard-edged target. A forward-peaked pattern means a long-range force. This is how Rutherford told a small, dense nucleus from a soft pudding.
The mass ratio is the neutron-moderation story. A neutron has mass $m$ and the nucleus has mass $M$, so $\mu=m/M\approx1/A$. For a heavy nucleus (large $A$) the neutron bounces back with almost all its energy and the nucleus hardly moves. This is the book case. For light nuclei such as hydrogen ($\mu$ near 1) the target recoils a lot, the neutron loses much of its energy, and the neutron can no longer scatter backwards in the lab. That is why water and carbon slow neutrons in a reactor. Raise $m/M$ in the simulation to watch this happen.
A limit of this model. Real particles are waves. When the de Broglie wavelength is much larger than the ball, hard-sphere scattering is still isotropic but the cross-section becomes $4\pi a^2$, four times the geometric area. At short wavelengths it tends to $2\pi a^2$. The classical answer $\pi a^2$ is the middle step, and holds when the wavelength is small compared with $a$ and the particle only bounces. (A comparison from a later theory, not part of the book problem.)
(iii-c) How to Read the Picture, the Beam Disc and the Three Graphs
- One particle or a beam? The normal film shows ONE particle with the aiming parameter $b$ you set. The button "Fire the beam" sends your $N$ particles one after another, each with its own random $b$. Each particle meets the ball at rest on its own, so they do not hit together. The graphs fill up as the particles are counted.
- Beam disc (top right). Look at the beam from the front. One dot is one particle, and its distance from the centre is its aiming parameter $b$. The dashed circle has radius $a=R+r$: dots inside hit, dots outside miss. The angle depends only on the distance $b$, so equal angles sit on circles around the centre (that is why you see rings). Dots near the centre are almost head-on and bounce back (orange). Dots near the edge are grazing and turn only a little (cyan). The faint dotted circle at $a/\sqrt2$ is the $90^\circ$ boundary. The amber ring holds the particles that end up inside your detector window. The white circle is the single particle in the film.
- Detector angle $\theta$ and opening $d\theta$. These are an ANGLE range of the deflection, not a speed range. Imagine a detector placed at angle $\theta$ from the original direction, opening $d\theta$ wide. $dP$ is the share of colliding particles that arrive in it. The amber wedge in the picture shows where it looks. In this elastic problem with a heavy ball, every particle keeps its speed $v$.
- Graph 1: across is the aiming parameter $b$, up is the deflection angle $\theta$ from $0^\circ$ to $180^\circ$. Grey dashed is the book case.
- Graph 2: across is the deflection angle $\theta$. Up is $dP/d\theta$, the share of colliding particles per radian of angle (the "top" number is the scale). Blue bars come from the sample, green is the theory $\tfrac12\sin\theta$. The area under all bars is 1.
- Graph 3: across is the angle $\theta$, up is the probability $P$ that a particle is deflected by less than $\theta$, from 0 to 1. At $90^\circ$ it reads the front-hemisphere probability.
- The green arrow at the collision. It is the push $\Delta v=2v\cos\varphi/(1+m/M)$ that the ball gives, along the line of centres. It changes with $b$, $R$, $r$, the speed and $m/M$. The small box at the lower left adds the vectors: $v$ before + $\Delta v$ = $v$ after.
(iv) The Solution Viewed in Each Frame
Numbers use $R=5$, $r=1$ (length units), $b=3$, and $\theta=120^\circ$ with a window $d\theta=5^\circ$. The live box above follows your sliders. The mass of the particle is $m$ and the ball's is $M$. We assume speeds are far below the speed of light, so ordinary momentum and energy rules are fine. The angle result does not depend on the speed.
1. Ground (Lab) Frame: the ball is at rest
First we find when the particle touches the ball, because that fixes the size of the target. The centres are $a=R+r$ apart at touch. In symbols $a=R+r$. With numbers $a=5+1$ length units. This gives $a=6$. Particles with $b>a$ miss and are not deflected.
We assume the ball is so heavy that it does not move ($m/M\to0$), and that the surfaces are smooth, so the push is along the line of centres. We also assume the collision is elastic, so the particle's speed stays $v$. These are used in the next paragraph.
Next we find the deflection angle, because that is part (a). At touch the particle's centre is at $(-\sqrt{a^2-b^2},\,b)$ when the ball's centre is the origin. The unit vector from the ball's centre to the particle's centre is $\hat n=(-\cos\varphi,\ \sin\varphi)$, where $\sin\varphi=\dfrac{b}{a}$ (a right triangle with sides $b$ and hypotenuse $a$). So $\varphi$ is the angle between the line of centres and the beam's path. With numbers $\sin\varphi=\dfrac{3}{6}$, giving $\varphi=30^\circ$.
With $\vec v_i=(v,0)$ we get $\vec v_i\cdot\hat n=-v\cos\varphi$. So $\vec v_f=\big(v-2v\cos^2\varphi,\ 2v\sin\varphi\cos\varphi\big)=v\,(-\cos2\varphi,\ \sin2\varphi)$.
A velocity of size $v$ at angle $\theta$ has components $v(\cos\theta,\sin\theta)$. Matching, $\cos\theta=-\cos2\varphi=\cos(\pi-2\varphi)$ and $\sin\theta=\sin2\varphi=\sin(\pi-2\varphi)$. So $\theta=\pi-2\varphi$.
Now we turn $\varphi$ into $b$. In symbols $b=a\sin\varphi=a\sin\!\big(\tfrac{\pi}{2}-\tfrac{\theta}{2}\big)=a\cos\tfrac{\theta}{2}$. With numbers $b=6\cos60^\circ=6\times0.5$. This gives $b=3$, which matches the start. So $\theta=2\arccos(b/a)$ for $b\le a$. At $b=0$, $\theta=\pi$ (bounces straight back). At $b=a$, $\theta=0$.
For part (b), we want the share of colliding particles in the angle window. The beam is uniform, so the share is area over area. The hitting particles fill a disk of area $\pi a^2$. Those with aiming parameter between $b$ and $b+db$ fill a ring of area $2\pi b\,|db|$. In symbols $dP=\dfrac{2\pi b\,|db|}{\pi a^2}$. We assume the fraction is counted among the particles that actually collide. From $b=a\cos\tfrac{\theta}{2}$ we get $db=-\tfrac{a}{2}\sin\tfrac{\theta}{2}\,d\theta$. Then $2\pi b\,|db|=2\pi\,a\cos\tfrac{\theta}{2}\cdot\tfrac{a}{2}\sin\tfrac{\theta}{2}\,d\theta=\tfrac{\pi a^2}{2}\sin\theta\,d\theta$, using $2\sin x\cos x=\sin2x$. So $dP=\tfrac12\sin\theta\,d\theta$. With numbers $\tfrac12\sin120^\circ\times0.0873=\tfrac12\times0.8660\times0.0873=0.0378$. This is about 3.8% of colliding particles. (The exact window is $\tfrac12[\cos\theta-\cos(\theta+d\theta)]=0.0368$, because a 5° window is not very thin.)
For part (c), we add up the shares for all $\theta$ below $\pi/2$. In symbols $P=\int_0^{\pi/2}\tfrac12\sin\theta\,d\theta=\tfrac12\big[-\cos\theta\big]_0^{\pi/2}$. With numbers $\tfrac12\big[-0-(-1)\big]$. This gives $P=\tfrac12$. It does not depend on $R$ or $r$, because the sizes cancel in the ratio of areas.
2. Centre-of-Mass Frame: total momentum is zero
We want the same results in the frame where the total momentum is zero. We assume the ball may have any mass ratio $\mu=m/M$ here, and we recover the book's case when $\mu\to0$. The frame moves at constant velocity, so it is inertial and needs no pseudo-force.
First the frame's speed. The total momentum in the ground frame is $mv$ and the total mass is $m+M$. In symbols $V=\dfrac{mv}{m+M}=\dfrac{\mu}{1+\mu}v$. With $\mu=0.5$: $V=\dfrac{0.5}{1.5}v=0.333v$. In this frame the particle moves at $u=v-V=\dfrac{v}{1+\mu}=0.667v$ and the ball at $-V$. Momenta: $mu=\dfrac{mv}{1+\mu}$ and $-MV=-\dfrac{mv}{1+\mu}$. They sum to zero.
The motion about the centre of mass is a mirror bounce of the relative velocity. The aiming parameter is the same $b$ in every frame, because it is a distance across the motion. So the same mirror-bounce box gives $\theta^*=\pi-2\varphi$, $\sin\varphi=b/a$, and $dP=\tfrac12\sin\theta^*\,d\theta^*$ exactly, for any $\mu$. Here $\theta^*$ is the deflection angle in this frame.
Now we turn to the ground frame, because the book's question is about it. After the bounce the particle's speed in the centre-of-mass frame is still $u$, turned by $\theta^*$. Adding $V$ back gives the ground-frame components $v_x=V+u\cos\theta^*$ and $v_y=u\sin\theta^*$. In symbols $\tan\theta=\dfrac{u\sin\theta^*}{V+u\cos\theta^*}=\dfrac{\sin\theta^*}{\mu+\cos\theta^*}$, since $V/u=\mu$. With $\mu=0.5$ and $\theta^*=120^\circ$: $\tan\theta=\dfrac{0.8660}{0.5-0.5}\to\infty$, so $\theta=90^\circ$ exactly. When $\mu\to0$, $\theta=\theta^*$ and the book's answers hold.
For part (c) in the ground frame, the particle moves forward when $\mu+\cos\theta^*>0$. In symbols $\theta^*<\arccos(-\mu)$. Then $P=\int_0^{\arccos(-\mu)}\tfrac12\sin\theta^*d\theta^*=\tfrac12\big[1-\cos\theta^*\big]=\dfrac{1+\mu}{2}$ for $\mu<1$ (and $P=1$ for $\mu\ge1$). With $\mu=0.5$: $P=\dfrac{1.5}{2}=0.75$. With $\mu\to0$: $P=\tfrac12$, the book's answer.
3. Particle Rest Frame: the particle starts at rest
We want to see the bounce as the particle's rider sees it. We assume the ball is heavy ($\mu\to0$), and speeds are small so plain velocity shifts work. The frame moves at $v$ along the beam, so the particle starts at rest and the ball rushes at it with velocity $-v$.
The only force is the ball's push on the particle along the line of centres $\hat n$. The relative speed along $\hat n$ is $v\cos\varphi$, and an elastic bounce flips it. In symbols the particle's velocity change is $\Delta\vec v=2v\cos\varphi\,\hat n$. With numbers $2v\cos30^\circ=1.732\,v$. So the particle leaves with speed $1.732\,v$ along $\hat n=(-\cos\varphi,\sin\varphi)$ pointing away from the ball's centre.
To compare with the book, we shift back to the ground frame by adding $v$ along $x$. The change of velocity does not depend on the frame, so $\vec v_f=(v,0)+2v\cos\varphi\,(-\cos\varphi,\sin\varphi)=v(-\cos2\varphi,\ \sin2\varphi)$. This is the same as in the boxed derivation. So $\theta=\pi-2\varphi=120^\circ$, $b=a\cos(\theta/2)$ and $dP=\tfrac12\sin\theta\,d\theta$.
In this frame the particle first stands still and then moves at an angle $\varphi$ from the line of centres. It is like the particle being kicked by a moving wall. The ball's speed hardly changes, because it is heavy.
- Changes: the starting velocities and the picture. In the centre-of-mass frame the deflection angle $\theta^*$ is bigger than the ground angle $\theta$ when $\mu>0$.
- Same (heavy ball): $\theta=\pi-2\varphi$, $b=a\cos(\theta/2)$, $dP=\tfrac12\sin\theta\,d\theta$ and $P=\tfrac12$, and the aiming parameter $b$ itself.
(v) The Solution by Each Method
Method 1: Momentum Method (impulse along the normal)
- We want the change of the particle's momentum. A smooth ball pushes only along $\hat n$, so the impulse $\vec J=J\hat n$ and the sideways momentum does not change.
- We assume an elastic bounce, so the speed along $\hat n$ flips: $J=2mv\cos\varphi$. In symbols $\Delta\vec p=2mv\cos\varphi\,\hat n$. With numbers (per unit $mv$) $2\cos30^\circ=1.732$.
- Adding to the start gives $\vec v_f=v(-\cos2\varphi,\sin2\varphi)$, so $\theta=\pi-2\varphi$ and $b=a\cos\frac{\theta}{2}$.
- The ring-area share then gives $dP=\tfrac12\sin\theta\,d\theta$ and $P=\tfrac12$ as shown in the ground frame.
Result: $\theta=2\arccos\frac{b}{R+r}$, $dP=\tfrac12\sin\theta\,d\theta$, $P=\tfrac12$.
Method 2: Energy Method
- We want the particle's final speed. Elastic means kinetic energy is conserved. The ball is heavy and stays at rest, so $\tfrac12mv_f^2=\tfrac12mv^2$ and $v_f=v$.
- Energy alone does not give direction. Add that the sideways velocity is unchanged (a smooth push is along $\hat n$). Then the normal part keeps its size, and since the particle must leave the ball, its sign flips.
- That is the mirror bounce, so $\theta=\pi-2\varphi$ with $\sin\varphi=b/a$. The distribution and $P=\tfrac12$ follow as before.
Method 3: Graph Method (areas on the beam picture)
- We want to read the probabilities from areas. The hitting particles fill a disk of radius $a$, area $\pi a^2$. The simulation plots $\theta$ against $b$, and the share against $\theta$.
- The front hemisphere ($\theta<\pi/2$) needs $b>a\cos45^\circ=a/\sqrt2$. In symbols $P=\dfrac{\pi a^2-\pi(a/\sqrt2)^2}{\pi a^2}=1-\tfrac12$. With numbers $\dfrac{36\pi-18\pi}{36\pi}$, which is $0.5$.
- The cumulative graph $\tfrac12(1-\cos\theta)$ passes through $0.5$ at $\theta=90^\circ$. The histogram of a random sample peaks at $\theta=90^\circ$ with height $0.5$ and its bars follow $\tfrac12\sin\theta$.
Method 4: Quantum-Rule Method
Not suited. The problem asks for classical elastic bounces of hard balls. (As a comparison from a later theory, waves would also spread by diffraction when their wavelength is not much smaller than the ball.)
Method 5: Level-Diagram Method
Does not apply. There are no energy levels in this problem.
Method 6: Check Method (units and limits)
- Units: angles are in radians and $dP$, $P$ are pure numbers, so no length units survive. That fits a share of particles.
- Head-on: $b=0$ gives $\theta=\pi$, a straight bounce back. Grazing: $b=a$ gives $\theta=0$. Both are right.
- Total: $\int_0^\pi\tfrac12\sin\theta\,d\theta=\tfrac12[-\cos\theta]_0^\pi=1$. All colliding particles are counted.
- Point particle: $r\to0$ gives $a=R$ and the same formulas. Size cancels, so $P=\tfrac12$ for any $R$, $r$.
Fastest: Method 3 for part (c), Method 1 for parts (a) and (b). Most reliable in an exam: Method 1 with Method 6 as a check. All give $\theta=2\arccos(b/a)$, $dP=\tfrac12\sin\theta\,d\theta$, $P=\tfrac12$.
(vi) What Each Slider Changes
| Slider | Outputs it changes | How | Because |
|---|---|---|---|
| Ball radius $R$ | $a$, $\varphi$, $\theta$, $\sigma=\pi a^2$, the picture | $a=R+r$; $\sin\varphi=b/a$ | A bigger target changes the angle for the same $b$. It does not change $dP$ or $P$. |
| Particle radius $r$ | same as $R$ | $a=R+r$ | Only the sum $R+r$ matters. |
| Aiming parameter $b$ | $\varphi$, $\theta$, picture | $\theta=2\arccos(b/a)$; none beyond $a$ | Aim decides the bounce angle. A miss gives $\theta=0$. |
| Angle $\theta$ (and inverse mode) | window centre, $dP$; $b$ in inverse mode | $dP=\tfrac12\sin\theta\,d\theta$ | Moves the amber window on the graphs, the amber ring in the beam disc and the amber wedge in the picture. It does not move the particle. |
| Window $d\theta$ | $dP$ | proportional (thin windows) | A wider window collects more particles. |
| Mass ratio $m/M$ | lab angle, front probability $P=(1+\mu)/2$, ball recoil | $\tan\theta=\sin\theta^*/(\mu+\cos\theta^*)$ | A lighter ball moves. The lab angle differs from the centre-of-mass angle, and the bars and curves change shape in the ground frame. The CM frame needs $\mu$ above about 0.05 to look different from the ground frame. |
| Sample size $N$ | number of dots in the beam disc, bar noise, the $\pm$ on the measured fractions | scatter $\propto1/\sqrt{\text{hits}}$ | It is a test of the theory. More particles make the measured bars agree better with the green curve. The theory does not change. |
| Particle speed, zoom | film speed, drawing size | none (display) | Hard-sphere angles do not depend on speed or on how large we draw. |
(vii) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(viii) Common Exam Traps
- Using $R$ instead of $R+r$. The particle's centre cannot get closer than $R+r$.
- Writing $\theta=2\varphi$. The turn is $\pi-2\varphi$, because $\varphi$ is measured from the line of centres.
- Forgetting the absolute value in $|db|$. The shares must be positive.
- Counting particles that miss. The fractions are among colliding particles only.
- Treating "front hemisphere" as the full $180^\circ$ or as $\theta>\pi/2$. It is $\theta<\pi/2$, and it gives $\tfrac12$.
Verified: live numbers, the three frames and the methods all agree for $R=5$, $r=1$, $b=3$: $\varphi=30^\circ$, $\theta=120^\circ$, $dP=0.0378$ for $d\theta=5^\circ$, $P=\tfrac12$ (heavy ball).