EXN.2.AHL.TZ0.7

pestleMathematicsAIHLPaper 2EXN· ahl-3-12-vector-applications-to-kinematicssource ↗

A ball is attached to the end of a string and spun horizontally. Its position relative to a given point, $O$ , at time $t$ seconds, $t \geq 0$ , is given by the equation

$r = (\begin{matrix} 1.5 \cos (0.1 t^{2}) \\ 1.5 \sin (0.1 t^{2}) \end{matrix})$ where all displacements are in metres.

The string breaks when the magnitude of the ball’s acceleration exceeds $20 {ms}^{- 2}$ .

Show that the ball is moving in a circle with its centre at $O$ and state the radius of the circle.

[4]

Find an expression for the velocity of the ball at time $t$ .

[2]

b.i.

Hence show that the velocity of the ball is always perpendicular to the position vector of the ball.

[2]

b.ii.

Find an expression for the acceleration of the ball at time $t$ .

[3]

c.i.

Find the value of $t$ at the instant the string breaks.

[3]

c.ii.

How many complete revolutions has the ball completed from $t = 0$ to the instant at which the string breaks?

[3]

c.iii.

Markscheme / solution

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

$|r| = \sqrt{1 . 5^{2} \cos^{2} (0.1 t^{2}) + 1 . 5^{2} \sin^{2} (0.1 t^{2})}$ M1

$= 1.5$ as $\sin^{2} θ + \cos^{2} θ = 1$ R1

Note: use of the identity needs to be explicitly stated.

Hence moves in a circle as displacement from a fixed point is constant. R1

Radius $= 1.5 m$ A1

[4 marks]

$v = (\begin{matrix} - 0.3 t \sin (0.1 t^{2}) \\ 0.3 t \cos (0.1 t^{2}) \end{matrix})$ M1A1

Note: M1 is for an attempt to differentiate each term

[2 marks]

b.i.

$v ∙ r = (\begin{matrix} 1.5 \cos (0.1 t^{2}) \\ 1.5 \sin (0.1 t^{2}) \end{matrix}) ∙ (\begin{matrix} - 0.3 t \sin (0.1 t^{2}) \\ 0.3 t \cos (0.1 t^{2}) \end{matrix})$ M1

Note: M1 is for an attempt to find $v ∙ r$

$= 1.5 \cos (0.1 t^{2}) \times (- 0.3 t \sin (0.1 t^{2})) + 1.5 \sin (0.1 t^{2}) \times 0.3 t \sin (0.1 t^{2}) = 0$ A1

Hence velocity and position vector are perpendicular. AG

[2 marks]

b.ii.

$a = (\begin{matrix} - 0.3 \sin (0.1 t^{2}) - 0.06 t^{2} \cos (0.1 t^{2}) \\ 0.3 \cos (0.1 t^{2}) - 0.06 t^{2} \sin (0.1 t^{2}) \end{matrix})$ M1A1A1

[3 marks]

c.i.

${(- 0.3 \sin (0.1 t^{2}) - 0.06 t^{2} \cos (0.1 t^{2}))}^{2} + {(0.3 \cos (0.1 t^{2}) - 0.06 t^{2} \sin (0.1 t^{2}))}^{2} = 400$ (M1)(A1)

Note: M1 is for an attempt to equate the magnitude of the acceleration to $20$ .

$t = 18.3 (18.256 \dots) (s)$ A1

[3 marks]

c.ii.

Angle turned through is $0.1 \times 18 . 256^{2} =$ M1

$= 33.329 \dots$ A1

$\frac{33.329}{2 π}$ M1

$\frac{33.329}{2 π} = 5.30 \dots$

$5$ complete revolutions A1

[4 marks]

c.iii.

Examiners’ report

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b.i.

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b.ii.

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c.i.

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c.ii.

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c.iii.