Further dynamics

WME03Topic 3.33 questions

Variable forces, Newton's law of gravitation and simple harmonic motion.

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Questions

Multiple choice: 1
Short answer: 1
Calculation: 1
Total: 3

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Sample questions

Original practice questions written in the style of the exam.

Question 16 marks
A particle $P$ moves with simple harmonic motion about a centre $O$ . The amplitude is $3\,\text{m}$ and the maximum speed of $P$ is $6\,\text{m s}^{-1}$ .

(a) Find the speed of $P$ when it is $2\,\text{m}$ from $O$ .

(b) Find the time taken for $P$ to move directly from $O$ to a point $1.5\,\text{m}$ from $O$ .
Mark scheme
- M1 $a\omega = 6$ with $a = 3$ (1)
- A1 $\omega = 2$ (1)
- M1 $v^{2} = ω^{2} ($ (1)
Question 25 marks
The Earth is modelled as a fixed sphere of radius $R$ . At a distance $x \geq R$ from the centre of the Earth, the gravitational force on a particle of mass $m$ has magnitude $\dfrac{mgR^2}{x^2}$ .

a^{2}

-

x^{2}

)

=

4

(

9

-

4

)

v^2 = \omega^2(a^2 - x^2) = 4(9 - 4)

v^{2} = ω^{2} (a^{2} - x^{2}) = 4 (9 - 4)

v = 2\sqrt{5} \approx 4.47\,\text{m s}^{-1}

(1)

M1 Uses

x = 3\sin 2t

(starting at

O

) with

x = 1.5

(1)

\sin 2t = \tfrac{1}{2} \Rightarrow t = \dfrac{\pi}{12} \approx 0.262\,\text{s}

(1)

x2mgR2

A particle is projected vertically upwards from the surface with speed $u$ and comes to instantaneous rest at a height $R$ above the surface. Ignoring air resistance, show that $u^2 = gR$ .

Mark scheme

B1 Equation of motion: $mv\dfrac{dv}{dx} = -\dfrac{mgR^2}{x^2}$ (1)
M1 Separates and integrates: $\tfrac{1}{2}v^2 = \dfrac{gR^2}{x} + c$ (1)
A1 $v = u$ at $x = R$ gives $c = \tfrac{1}{2}u^2 - gR$ (1)
M1 Sets $v = 0$ at $x = 2R$ : $0 = \dfrac{gR}{2} + \tfrac{1}{2}u^2 - gR$ (1)
A1 $u^2 = gR$ convincingly shown (1)

gR2

−

21u2−