CAIE A-Level Further Math 3.3 Circular motion Question Bank

Angular speeds of P and Q are equal,
so $\frac{\nu_{Q}}{x}=\frac{\nu_{P}}{3 a-x}$

M1

$v_{Q}=\frac{2 x \sqrt{a g}}{3 a-x}$

A1

Shown convincingly: angular speeds equal stated.
AG

2

For $P: T+m g \cos 60^{\circ}=\frac{m \times 4 a g}{3 a-x}$

B1

For $Q: T-m g \cos 60^{\circ}=\frac{m v_{Q}^{2}}{x}$

B1

Eliminate $T:-m g \cos 60^{\circ}+\frac{m \cdot 4 a g}{3 a-x}=m g \cos 60^{\circ}+\frac{m v_{Q}^{2}}{x}$

M1

$$\begin{aligned} \frac{m \times 4 a g}{3 a-x}=1+\frac{m x 4 a g}{(3 a-x)^{2}} 4 a(3 a-x)=(3 a-x)^{2}+4 a x, x^{2}+2 a x-3 a^{2}=0 \end{aligned}$$

M1

Solve to find x.
Obtain 3-term quadratic equation.

$(x-a)(x+3 a)=0, \quad x=a$

A1

5

B1

KEs correct.

Energy changes from initial position:

Gain in KE of $P: \frac{1}{2} m\left(4 a g-u^{2}\right)$

Loss in KE of $Q: \frac{1}{2} m\left(\left(\frac{u}{2}\right)^{2}-v_{Q}^{2}\right)$

Loss in GPE of $P=m g(3 a-x)\left(1-\cos 60^{\circ}\right)(=m g a)$

Gain in GPE of $Q=m g x\left(1-\cos 60^{\circ}\right)\left(=\frac{1}{2} m g a\right)$

B1FT

GPEs correct.

$\frac{1}{2} m\left(4 a g-u^{2}\right)-\frac{1}{2} m\left(\left(\frac{u}{2}\right)^{2}-v_{Q}^{2}\right)=-m g x\left(1-\cos 60^{\circ}\right)+m g(3 a-x)\left(1-\cos 60^{\circ}\right)$

M1

Energy equation.

Simplify: $4 a g-\frac{5}{4} u^{2}+a g=a g$$$u^{2}=\frac{16}{5} a g, \quad u=\frac{4}{5} \sqrt{5 a g}$$

A1

AEF

4