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2019 Paper 1 Q10
In this question, the \(x\)-axis is horizontal and the positive \(y\)-axis is vertically upwards. A particle is projected from the origin with speed \(u\) at an angle \(\alpha\) to the vertical. The particle passes through the fixed point \((h \tan \beta, h)\), where \(0 < \beta < 90^{\circ}\) and \(h > 0\).
- Show that
$$c^2 - ck \cot \beta + 1 + k \cot^2 \beta = 0, \quad (*)$$
where \(c = \cot \alpha\) and \(k = \frac{2u^2}{gh}\).
You are given that there are two distinct values of \(\alpha\) that satisfy equation \((*)\). Let \(\alpha_1\) and \(\alpha_2\) be these values.
- Show that $$\cot \alpha_1 + \cot \alpha_2 = k \cot \beta.$$ Show also that $$\alpha_1 + \alpha_2 = \beta.$$
- Show that $$k > 2(1 + \sec \beta).$$
- By considering the greatest height attained by the particle, show that \(k \geq 4 \sec^2 \alpha\).
Solution:
- The horizontal position of the particle at time \(t\) is \(u \sin\alpha t\), so \(T = \frac{h \tan \beta}{u \sin \alpha}\)
The vertical position of the particle at this time \(T\) satisifes:
\begin{align*}
&& h &= u \cos\alpha \frac{h \tan \beta}{u \sin\alpha} - \frac12 g \left ( \frac{h \tan \beta}{u \sin\alpha} \right)^2 \\
&&&= h\cot \alpha \tan \beta - \frac{gh^2}{2u^2} \tan^2 \beta \cosec^2 \beta \\
\Rightarrow && 1 &= c \tan \beta - \frac{1}{k} \tan^2 \beta (1 + c^2) \\
\Rightarrow && k \cot^2 \beta &= kc\cot \beta -1-c^2 \\
\Rightarrow && 0 &= c^2 -ck \cot \beta + 1 + k \cot^2 \beta
\end{align*}
- As a quadratic in \(c\) the sum of the roots is \(k \cot \beta\), therefore \(\cot \alpha_1 + \cot \alpha_2 = k \cot \beta\). We also have that \(\cot \alpha_1 \cot \alpha_2 = 1 + k \cot^2 \beta\), so \begin{align*} && \cot (\alpha_1 + \alpha_2) &= \frac{\cot \alpha_1 \cot \alpha_2-1}{\cot \alpha_1 + \cot \alpha_2} \\ &&&= \frac{1 + k \cot^2 \beta - 1}{k \cot \beta} \\ &&&= \cot \beta \\ \Rightarrow && \beta &= \alpha_1 + \alpha_2 \pmod{\pi} \end{align*} but since \(\alpha_i \in (0, \frac{\pi}{2})\) the equation must hold exactly.
- Since it has two real roots we must have \begin{align*} && 0 &<\Delta = k^2 \cot^2 \beta - 4 (1 + k \cot^2 \beta) \\ &&&= k^2 \cot^2 \beta-4k \cot^2 \beta -4 \\ &&&= \cot^2 \beta (k^2 - 4k - 4(\sec^2 \beta - 1)) \\ &&&= \cot^2 \beta ( (k-2)^2 -4\sec^2 \beta) \\ \Rightarrow && k &> 2 + 2\sec \beta = 2(1+\sec \beta) \end{align*}
- The greatest height will satisfy \(v^2 = u^2 + 2as\) so \(0 = u^2 \cos^2 \alpha - 2gh_{max} \Rightarrow 4\sec^2 \alpha = \frac{2u^2}{gh_{max}} = k_{max}\), but this decreases with \(h\), so the smallest \(k\) can be is \(4\sec^2 \alpha\), ie \(k \geq 4 \sec^2 \alpha\)
2013 Paper 1 Q9
Two particles, \(A\) and \(B\), are projected simultaneously towards each other from two points which are a distance \(d\) apart in a horizontal plane. Particle \(A\) has mass \(m\) and is projected at speed \(u\) at angle \(\alpha\) above the horizontal. Particle \(B\) has mass \(M\) and is projected at speed \(v\) at angle \(\beta\) above the horizontal. The trajectories of the two particles lie in the same vertical plane. The particles collide directly when each is at its point of greatest height above the plane. Given that both \(A\) and \(B\) return to their starting points, and that momentum is conserved in the collision, show that \[ m\cot \alpha = M \cot \beta\,. \] Show further that the collision occurs at a point which is a horizontal distance \(b\) from the point of projection of \(A\) where \[ b= \frac{Md}{m+M}\, , \] and find, in terms of \(b\) and \(\alpha\), the height above the horizontal plane at which the collision occurs.
Solution: Since \(A\) and \(B\) return to their starting points, and at their highest points there is no vertical component to their velocities, their horizontal must perfectly reverse, ie \begin{align*} && m u \cos \alpha - M v \cos \beta &= -m u \cos \alpha + M v \cos \beta \\ \Rightarrow && mu \cos \alpha &= Mv \cos \beta \end{align*} Since they reach their highest points at the same time, they must have the same initial vertical speed, ie \(u \sin \alpha = v \sin \beta\), so \begin{align*} && m v \frac{\sin \beta}{\sin \alpha} \cos \alpha &= M v \cos \beta \\ \Rightarrow && m \cot \alpha &= M \cot \beta \end{align*} The horizontal distance travelled by \(A\) & \(B\) will be: \begin{align*} && d_A &= u \cos \alpha t \\ && d_B &= v \cos \beta t \\ \Rightarrow && \frac{d_A}{d_A+d_B} &= \frac{u \cos \alpha}{u \cos \alpha + v \cos \beta} \\ &&&= \frac{\frac{M}{m}v \cos \beta}{\frac{M}{m}v \cos \beta + v \cos \beta} \\ &&&= \frac{M}{M+m} \\ \Rightarrow && d_A = b &= \frac{Md}{m+M} \end{align*} Applying \(v^2 = u^2 + 2as\) we see that \begin{align*} && 0 &= u \sin \alpha - gt \\ \Rightarrow && t &= \frac{u \sin \alpha}{g} \\ && b &=u \cos \alpha \frac{u \sin \alpha}{g} \\ \Rightarrow && u^2 &= \frac{2bg}{\sin 2 \alpha} \\ && 0 &= u^2 \sin^2 \alpha - 2g h \\ \Rightarrow && h &= \frac{u^2 \sin^2 \alpha}{2g} \\ &&&= \frac{2bg}{\sin 2 \alpha} \frac{ \sin^2 \alpha}{2g} \\ &&&= \frac12 b \tan \alpha \end{align*}
2004 Paper 1 Q9
A particle is projected over level ground with a speed \(u\) at an angle \(\theta\) above the horizontal. Derive an expression for the greatest height of the particle in terms of \(u\), \(\theta\) and \(g\). A particle is projected from the floor of a horizontal tunnel of height \({9\over 10}d\). Point \(P\) is \({1\over 2}d\) metres vertically and \(d\) metres horizontally along the tunnel from the point of projection. The particle passes through point \(P\) and lands inside the tunnel without hitting the roof. Show that \[ \arctan \textstyle {3 \over 5} < \theta < \arctan \, 3 \;. \]
Solution: \begin{align*} && v^2 &= u^2 + 2as \\ (\uparrow): && 0 &= (u \sin \theta)^2 - 2gh \\ \Rightarrow && h &= \frac{u^2 \sin^2 \theta}{2g} \end{align*} To avoid hitting the ceiling \begin{align*} && \frac9{10}d &>\frac{u^2 \sin^2 \theta}{2g} \\ \Rightarrow && \frac{gd}{u^2}&>\frac{5\sin^2 \theta}{9} \\ \end{align*} In order to pass through \(P\) we need \begin{align*} && d &= u \cos \theta t \\ && \frac12 d &= u \sin \theta t - \frac12 g t^2 \\ \Rightarrow && \frac12 &= \tan \theta - \frac12 \frac{g d}{u^2} \sec^2 \theta \\ &&&<\tan \theta - \frac12 \frac{5 \sin^2 \theta}{9\cos^2 \theta} \\ \Rightarrow && 0 & >5 \tan^2 \theta -18\tan \theta +9 \\ &&&= (5\tan \theta - 3)(\tan \theta - 3) \\ \\ \Rightarrow && \tan \theta &\in \left ( \frac35, 3\right) \\ && \theta &= \left (\arctan \tfrac35, \arctan 3 \right) \end{align*}
1998 Paper 1 Q10
A shell explodes on the surface of horizontal ground. Earth is scattered in all directions with varying velocities. Show that particles of earth with initial speed \(v\) landing a distance \(r\) from the centre of explosion will do so at times \(t\) given by \[ {\textstyle \frac{1}{2}} g^2t^2=v^{2}\pm\surd(v^{4}-g^{2}r^{2}). \] Find an expression in terms of \(v\), \(r\) and \(g\) for the greatest height reached by such particles.
1996 Paper 2 Q10
The plot of `Rhode Island Red and the Henhouse of Doom' calls for the heroine to cling on to the circumference of a fairground wheel of radius \(a\) rotating with constant angular velocity \(\omega\) about its horizontal axis and then let go. Let \(\omega_{0}\) be the largest value of \(\omega\) for which it is not possible for her subsequent path to carry her higher than the top of the wheel. Find \(\omega_{0}\) in terms of \(a\) and \(g\). If \(\omega>\omega_{0}\) show that the greatest height above the top of the wheel to which she can rise is \[\frac{a}{2}\left(\frac{\omega}{\omega_{0}} -\frac{\omega_{0}}{\omega}\right)^{\!\!2}.\]
Solution:
