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2022 Paper 2 Q10
D: 1500.0 B: 1500.0

  1. Show that, if a particle is projected at an angle \(\alpha\) above the horizontal with speed \(u\), it will reach height \(h\) at a horizontal distance \(s\) from the point of projection where \[h = s\tan\alpha - \frac{gs^2}{2u^2\cos^2\alpha}\,.\]
The remainder of this question uses axes with the \(x\)- and \(y\)-axes horizontal and the \(z\)-axis vertically upwards. The ground is a sloping plane with equation \(z = y\tan\theta\) and a road runs along the \(x\)-axis. A cannon, which may have any angle of inclination and be pointed in any direction, fires projectiles from ground level with speed \(u\). Initially, the cannon is placed at the origin.
  1. Let a point \(P\) on the plane have coordinates \((x,\, y,\, y\tan\theta)\). Show that the condition for it to be possible for a projectile from the cannon to land at point \(P\) is \[x^2 + \left(y + \frac{u^2\tan\theta}{g}\right)^2 \leqslant \frac{u^4\sec^2\theta}{g^2}\,.\]
  2. Show that the furthest point directly up the plane that can be reached by a projectile from the cannon is a distance \[\frac{u^2}{g(1+\sin\theta)}\] from the cannon. How far from the cannon is the furthest point directly down the plane that can be reached by a projectile from it?
  3. Find the length of road which can be reached by projectiles from the cannon. The cannon is now moved to a point on the plane vertically above the \(y\)-axis, and a distance \(r\) from the road. Find the value of \(r\) which maximises the length of road which can be reached by projectiles from the cannon. What is this maximum length?

2011 Paper 2 Q12
D: 1600.0 B: 1484.0

Xavier and Younis are playing a match. The match consists of a series of games and each game consists of three points. Xavier has probability \(p\) and Younis has probability \(1-p\) of winning the first point of any game. In the second and third points of each game, the player who won the previous point has probability \(p\) and the player who lost the previous point has probability \(1-p\) of winning the point. If a player wins two consecutive points in a single game, the match ends and that player has won; otherwise the match continues with another game.

  1. Let \(w\) be the probability that Younis wins the match. Show that, for \(p\ne0\), \[ w = \frac{1-p^2}{2-p}. \] Show that \(w>\frac12\) if \(p<\frac12\), and \(w<\frac12\) if \(p>\frac12\). Does \(w\) increase whenever \(p\) decreases?
  2. If Xavier wins the match, Younis gives him \(\pounds1\); if Younis wins the match, Xavier gives him \(\pounds k\). Find the value of \(k\) for which the game is `fair' in the case when \(p =\frac23\).
  3. What happens when \(p = 0\)?


Solution:

  1. We can be in several states.
    1. No points played
    2. Y just won the last point
    3. X just won the last point
    4. Y won the game
    5. X won the game
    The probability \(Y\) wins from any of these states are: \begin{align*} &&P_{-} &= p P_X + (1-p) P_Y &= w \\ &&P_Y &= p + (1-p)P_X \\ &&P_X &= (1-p)P_Y \\ \\ \Rightarrow &&& \begin{cases} P_Y - (1-p)P_X &= p \\ (1-p)P_Y -P_X &= 0\end{cases} \\ \Rightarrow && P_Y &= \frac{1}{1-(1-p)^2} \cdot p \\ &&&= \frac{1}{2-p} \\ && P_X &= \frac{1-p}{2-p} \\ && w &= \frac{p(1-p) + (1-p)}{2-p} \\ &&&= \frac{1-p^2}{2-p} \end{align*}
  2. If \(p = \frac23\) then \(w = \frac{1-\frac49}{2-\frac23} = \frac{5}{12}\). The game is fair if \(\mathbb{E}(result) = 0\), ie \(\frac{5}{12} \cdot k - \frac{7}{12} 1 \Rightarrow k = \frac{7}{5} = 1.4\)
  3. If \(p = 0\) then they will keep playing forever, since no-one can win two points in a row.