Problem 60

Problem 60: Newton’s cradle ( $✓$ $✓$ ) 1999 Paper II

$N$ particles $P_{1}$ , $P_{2}$ , $P_{3}$ , $\dots$ , $P_{N}$ with masses $m$ , $q m$ , $q^{2} m$ , $\dots$ , $q^{N - 1} m$ , respectively, are at rest at distinct points along a straight line in gravity-free space. The particle $P_{1}$ is set in motion towards $P_{2}$ with velocity $V$ and in every subsequent impact the coefficient of restitution is $e$ , where $0 < e < 1$ .

(i): Show that after the ﬁrst impact the velocities of $P_{1}$ and $P_{2}$ are $(\frac{1 - e q}{1 + q}) V and (\frac{1 + e}{1 + q}) V,$
respectively.
(ii): Show that, if $q \leq e$ , then there are exactly $N - 1$ impacts.
(iii): Show further that, if $q = e$ , then the total loss of kinetic energy after all impacts have occurred is $\frac{1}{2} m e (1 - e^{N - 1}) V^{2} .$

Comments

This situation models the toy called ‘Newton’s Cradle’ which consists of four or more heavy metal balls suspended from a frame so that they can swing. At rest, they are in contact in a line. When the ﬁrst ball is raised and let swing, there follows a rather pleasing pattern of impacts. In this case, the coefficient of restitution is nearly 1 and the balls all have the same mass, so, as the ﬁrst displayed formula shows, the impacting ball is reduced to rest by the impact. At the ﬁrst swing of the ball, nothing happens except that the ﬁrst ball is reduced to rest and the last ball swings away. Note that this is consistent with the balls being separated by a very small amount; what actually happens is that the ball undergoes a small elastic deformation at the impact, and the impulse takes a small amount of time to be transmitted across the ball to the next ball.

Solution to problem 60

(i) Let $V_{1}$ and $V_{2}$ be the velocities of $P_{1}$ and $P_{2}$ after the ﬁrst collision. Using conservation of momentum and Newton’s law of impact at the ﬁrst collision results in the equations

m V = m V_{1} + m q V_{2}, V_{1} - V_{2} = - e V,

V_{1} = (\frac{1 - e q}{1 + q}) V and V_{2} = (\frac{1 + e}{1 + q}) V .

(ii) Note that $V_{1}$ is positive since $e q \leq e^{2} < 1$ . Conserving momentum and using Newton’s law (and doing a bit of algebra) shows that the speeds ${\hat{V}}_{2}$ and $V_{3}$ of $P_{2}$ and $P_{3}$ after the next collision are given by

{\hat{V}}_{2} = (\frac{1 - e q}{1 + q}) V_{2} = (\frac{1 - e q}{1 + q}) (\frac{1 + e}{1 + q}) V and V_{3} = (\frac{1 + e}{1 + q}) V_{2} = {(\frac{1 + e}{1 + q})}^{2} V .

(

*

)

Note that ${\hat{V}}_{2} \geq V_{1}$ so there is no further collision between $P_{1}$ and $P_{2}$ . Applying this argument at each collision shows that there are exactly $N - 1$ collisions: $P_{1}$ with $P_{2}$ ; $P_{2}$ with $P_{3}$ ; etc.

(iii) The speed of $P_{k}$ after it has hit $P_{k + 1}$ is (by extending $(*)$ )

(\frac{1 - e q}{1 + q}) {(\frac{1 + e}{1 + q})}^{k - 1} V,

and the speed of $P_{k + 1}$ after this collision and before it hits $P_{k + 2}$ is

{(\frac{1 + e}{1 + q})}^{k} V .

If $q = e$ ,

V_{1} = (\frac{1 - e^{2}}{1 + e}) V = (1 - e) V

and similarly the ﬁnal speeds of $P_{2}$ , $\dots$ , $P_{N - 1}$ are all $(1 - e) V$ . The ﬁnal speed of $P_{N}$ is $V$ . Thus the ﬁnal total kinetic energy is

\begin{array}{rcl} \frac{1}{2} (m + m q + \dots + m q^{N - 2}) {[(1 - e) V]}^{2} + \frac{1}{2} m q^{N - 1} V^{2} & = & \frac{1}{2} m \frac{1 - e^{N - 1}}{1 - e} {(1 - e)}^{2} V^{2} + \frac{1}{2} m e^{N - 1} V^{2} \\ = & \frac{1}{2} m (1 - e + e^{N}) V^{2}, \end{array}

(replacing all the $q$ ’s with $e$ ’s and summing the geometric progression). Thus the loss of kinetic energy is

\frac{1}{2} m V^{2} - \frac{1}{2} m (1 - e + e^{N}) V^{2},

as required.

Post-mortem

There are two tricky aspects to these multiple collision questions. First there is the matter of notation. Above, I have used a hat for the second collision of a particle ( ${\hat{V}}_{2}$ ), retaining the subscript for labelling particles. That works, but if $P_{2}$ undergoes another collision, entailing a double hat, it starts getting messy. You could use a different letter, but that gets confusing. The only good method is to use a double subscript: $V_{m, n}$ is the velocity of the $m$ th particle after the $n$ th collision. But that is a sledge hammer for this nut of a question.

The other tricky aspect is getting the signs right. I always think of velocity (not speed), and it is always positive for particles travelling to the right. I also use common sense to check each equation! As usual, a good diagram of each collision is essential.