Solved: A ball is attached to the end of a string and spun horizontally. Its position... [IB Mathematics Applications & Interpretation (Math AI)]

Question

HLPaper 2

A ball is attached to the end of a string and spun horizontally. Its position relative to a givenpoint, $O$ , 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 ofthe circle.

[4]

Verified

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]

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

[2]

Verified

Solution

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

Note: M1 is for an attempt to differentiate each term

[2 marks]

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

[2]

Verified

Solution

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

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

[3]

Verified

Solution

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

[3 marks]

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

[3]

Verified

Solution

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

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

[3]

Verified

Solution

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]

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