M3 June 2015 Q7
7. A solid smooth sphere, with centre \(O\) and radius \(r\), is fixed to a point \(A\) on a horizontal floor. A particle \(P\) is placed on the surface of the sphere at the point \(B\), where \(B\) is vertically above \(A\). The particle is projected horizontally from \(B\) with speed \(\dfrac{\sqrt{gr}}{2}\) and starts to move on the surface of the sphere. When \(OP\) makes an angle \(\theta\) with the upward vertical and \(P\) remains in contact with the sphere, the speed of \(P\) is \(v\).
The particle leaves the surface of the sphere when \(\theta = \alpha\).
After leaving the surface of the sphere, \(P\) moves freely under gravity and hits the floor at the point \(C\).
Given that \(r = 0.5\) m,
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m\dfrac{rg}{4} = mgr(1 - \cos\theta)\) | M1A1A1 |
| \(v^2 = \dfrac{rg}{4}(9 - 8\cos\theta)\) * | A1 |
| (4) |
Notes
M1 Attempting an energy equation. 2KE terms needed and a PE term.
Award if mass missing throughout, but not for use of \(v^2 = u^2 + 2as\)
A1 KE terms correct (and subtracted) Mass not needed if M mark earned
A1 PE correct Again, mass not needed if M mark earned
A1cso Obtaining the GIVEN answer
| Scheme | Marks |
|---|---|
| \((R)\ \ + mg\cos\theta = \dfrac{mv^2}{r}\) | M1A1 |
| \(R = 0 \quad mg\cos\alpha = \dfrac{mg}{4}(9 - 8\cos\alpha)\) | DM1 |
| \(12\cos\alpha = 9\) | |
| \(\cos\alpha = \dfrac{3}{4}\) or 0.75 | A1 |
| (4) |
Notes
M1 Attempting an equation of motion along the radius. Accel in either form, \((\pm)R\) may be included.
A1 Correct equation, with or without \((\pm)R\)
DM1 Set \(R = 0\) and substitute for \(v\)
A1 \(\cos\alpha = 3/4\) obtained
| Scheme | Marks |
|---|---|
| Initial vert comp of speed \(= \sqrt{\dfrac{3g}{8}}\sin\alpha = \sqrt{\dfrac{3g}{8}} \times \dfrac{\sqrt{7}}{4}\ \ (= 1.2679\ldots)\) | M1A1 |
| \(\dfrac{7}{8} = 1.2679\ldots t + \dfrac{1}{2}gt^2\) | M1 |
| \(7 = 10.143\ldots t + 39.2t^2\) | |
| \(39.2t^2 + 10.143\ldots t - 7 = 0\) | |
| \(t = \dfrac{-10.143 \pm \sqrt{10.143^2 + 4 \times 7 \times 39.2}}{2 \times 39.2}\) | DM1 |
| \(t = 0.3125\ldots\) | A1 |
| Horiz speed \(= \sqrt{\dfrac{3g}{8}}\cos\alpha = \dfrac{1}{4}\sqrt{\dfrac{27g}{8}}\) | |
| \(AC = \dfrac{1}{4}\sqrt{\dfrac{27g}{8}} \times 0.3125 + r\sin\alpha = 0.4493 + 0.3307 = 0.78\) m | M1A1cso |
| (7) | |
| (15 marks) |
Notes
M1 Attempting the initial vertical component of the speed
A1 Correct vertical component - decimal or exact
M1 Using \(s = ut + \tfrac{1}{2}at^2\) to form a quadratic in \(t\), with their vertical speed and attempt at the vertical distance Must satisfy \(0.5 < \text{distance} < 1\)
DM1 Solving their quadratic; formula must be shown (and correct) if answer is incorrect, but allow with \(+\sqrt{\ldots}\) instead of \(\pm\sqrt{\ldots}\)
A1 Correct \(t\). Give by implication if stored on a calculator and final answer correct
Second solution need not be shown; ignore any shown
M1 Using the horizontal speed and completing to obtain the required distance.
A1 \(AC = 0.78\) must be 2 sf.
ALT for (c):
M1A1 As main method above
M1 Use the horizontal speed and distance travelled as a projectile to get an expression for \(t\) and substitute in \(s = ut + \tfrac{1}{2}at^2\) Vertical distance must be between 0.5 and 1
DM1 Solve their quadratic - see above
A1 Correct (projectile) distance
M1A1 As main method above
7(c) Using energy etc:
M1 Using energy to get the speed at the floor. Can be from the top or the point of leaving the surface
A1 Correct speed at floor
M1 Using the horizontal component of the speed and Pythagoras to obtain the vertical component at the floor
M1 Using \(v = u + at\) vertically to get \(t\)
A1 Correct \(t\)
M1A1 Complete as main method