2020HSCShort answer

Q16 (a) (ii)

3 marks

Question

Given that $v<\dfrac{gm}{k}$ , show that when
$t=\frac{3m}{k}\ln2,$
the velocity of the larger mass is
$\frac{gm}{2k}.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Obtains $t=\frac{3m}{k}\ln\left(\frac{gm}{gm-kv}\right)$ , or equivalent merit	2
Integrates $\frac{dt}{dv}$ , or equivalent merit	1

Sample answer:

From part (i),

\frac{dv}{dt}=\frac{gm-kv}{3m}

\therefore t=\int_0^v\frac{3m}{gm-kv}\,dv

=-\frac{3m}{k}\ln(gm-kv)\Big|_0^v

=-\frac{3m}{k}[\ln(gm-kv)-\ln(gm)]

=\frac{3m}{k}\ln\left(\frac{gm}{gm-kv}\right)

When $t=\frac{3m}{k}\ln2$ , we have

\frac{gm}{gm-kv}=2

\Rightarrow1-\frac{kv}{gm}=\frac12

\Rightarrow v=\frac{gm}{2k}

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Syllabus alignment

Mechanics

Resisted motionDerive and solve motion models with resistance on horizontal and vertical paths, including terminal velocity.

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