Two particles move on a smooth horizontal table and collide. The masses of the particles are \(m\) and \(M\). Their velocities before the collision are \(u{\bf i}\) and \(v{\bf i}\,\), respectively, where \(\bf i\) is a unit vector and \(u>v\).
Their velocities after the collision are \(p{\bf i}\) and \(q{\bf i}\,\), respectively.
The coefficient of restitution between the two particles is \(e\), where \(e<1\).
Show that the loss of kinetic energy due to the collision is
\[
\tfrac12 m (u-p)(u-v)(1-e)\,,
\]
and deduce that \(u\ge p\).
Given that each particle loses the same (non-zero) amount of kinetic energy in the collision, show that
\[
u+v+p+q=0\,,
\]
and that, if \(m\ne M\),
\[
e= \frac{(M+3m)u + (3M+m)v}{(M-m)(u-v)}\,.
\]
Solution:
\begin{align*}
\text{COM}: && mu + Mv &= mp + Mq \\
\Rightarrow && m(u-p) &= M(q-v) \\
\text{NEL}: && q-p &= e(u-v) \\
&& q +ev &= p+eu
\\
&& \Delta \text{ k.e.} &= \frac12 m u^2 + \frac12 M v^2 -\frac12 m p^2 - \frac12 M q^2 \\
&&&= \frac12m (u^2 - p^2)+\frac12M(v^2-q^2) \\
&&&= \frac12m (u^2 - p^2)+\frac12M(v-q)(v+q) \\
&&&= \frac12m(u^2-p^2) - \frac12 m(u-p)(v+q) \\
&&&= \frac12 m(u-p) \left ( u+p-v-q\right) \\
&&&= \frac12 m(u-p) \left (u-v+(p-q)\right) \\
&&&= \frac12 m(u-p) \left (u-v-e(u-v)\right) \\
&&&= \frac12m(u-p)(u-v)(1-e)
\end{align*}
Since the loss in energy is positive, and \(m\), \(u-v\) and \(1-e\) are all positive, so is \(u-p\), ie \(u \geq p\)
Prove that, for any real numbers \(x\) and \(y\), \(x^2+y^2\ge2xy\,\).
Carol has two bags of sweets. The first bag contains \(a\) red sweets
and \(b\) blue sweets, whereas the second bag contains \(b\) red sweets and
\(a\) blue sweets, where \(a\) and \(b\) are positive integers. Carol shakes
the bags and picks
one sweet from each bag without looking. Prove that the probability
that
the sweets are of the same colour
cannot exceed the probability that
they are of different colours.
Simon has three bags of sweets. The first bag
contains
\(a\) red sweets, \(b\) white sweets and \(c\) yellow sweets, where \(a\), \(b\) and
\(c\) are positive integers. The second
bag contains
\(b\) red sweets, \(c\) white sweets and \(a\) yellow sweets. The third
bag contains
\(c\) red sweets, \(a\) white sweets and \(b\) yellow sweets.
Simon shakes the bags and
picks one sweet from each bag without looking.
Show that the probability that exactly two of the sweets are of the
same colour is
\[
\frac {3(a^2b+b^2c+c^2a+ab^2 + bc^2 +ca^2)}{(a+b+c)^3}\,,
\]
and find the probability that the sweets are all of the same colour.
Deduce that the probability that exactly two of
the sweets are of the same colour is at least 6 times the probability
that the sweets are all of the same colour.
I seat \(n\) boys and \(3\) girls in a line at random, so
that each order of the \(n+3\) children
is as likely to occur as any other. Let \(K\) be the maximum number of
consecutive girls in the line so, for example, \(K=1\) if
there is at least one boy between
each pair of girls.
Find \(\P(K=3)\).
Show that
\[\P(K=1)=
\frac{n(n-1)}{(n+2)(n+3)}\,.
\]
Find \(\E(K)\).
Solution:
If all the girls are say together there are \(n+1\) ways to place the block of 3 girls. There are \(\binom{n+3}{3}\) ways to choose where to place the girls in total, therefore:
\begin{align*}
&& \mathbb{P}(K =3) &= \frac{n+1}{\binom{n+3}3} \\
&&&= \frac{6(n+1)}{(n+3)(n+2)(n+1)} \\
&&&= \frac{6}{(n+3)(n+2)}
\end{align*}
If \(K= 1\) then all of the girls are separated. We can place three girls and two boys separating them, then we are allocating \(N-2\) boys to \(4\) gaps, ie \(\binom{N-2+3}{3} = \binom{N+1}{3}\).
\begin{align*}
&& \mathbb{P}(K=3) &= \frac{\binom{n+1}{3}}{\binom{n+3}{3}} \\
&&&= \frac{(n+1)n(n-1)}{(n+3)(n+2)(n+1)} \\
&&&= \frac{n(n-1)}{(n+3)(n+2)}
\end{align*}