2023TrialShort answer

Q15 (a) (i)

3 marks

Question

A particle of unit mass is moving in horizontal motion, subject to a resistance force of $v^2+v^3$ , where $v$ is the object’s velocity. The particle has initial velocity $v_0$ , where $v_0>0$ .

Find the distance $s$ travelled by the particle when its velocity is

v=\frac{v_0}{2}.

Try the question before revealing the answer.

Solution

ma=-(v^2+v^3)\qquad\text{(Newton's 2nd law)}

m=1\Rightarrow a=-(v^2+v^3)

v\frac{dv}{dx}=-(v^2+v^3)

\frac{dv}{dx}=-(v+v^2)

=-v(1+v)

\int_{v_0}^{v}\frac{dv}{v(1+v)}=-\int_0^x dx

\int_{v_0}^{v}\left(\frac1v-\frac1{v+1}\right)dv=-x

-x=\left[\ln|v|-\ln|v+1|\right]_{v_0}^{v}

=\left[\ln\left|\frac{v}{v+1}\right|\right]_{v_0}^{v}

=\left[\ln\left(\frac{v}{v+1}\right)\right]_{v_0}^{v}\qquad\text{(Since }v>0\text{)}

=\ln\left(\frac{v}{v+1}\right)-\ln\left(\frac{v_0}{v_0+1}\right)

=\ln\left(\frac{v(v_0+1)}{v_0(v+1)}\right)

x=\ln\left(\frac{v_0(v+1)}{v(v_0+1)}\right)

v=\frac{v_0}{2},\quad x=s

s=\ln\left(\frac{v_0\left(\frac{v_0}{2}+1\right)}{\frac{v_0}{2}(v_0+1)}\right)

=\ln\left(\frac{v_0+2}{v_0+1}\right)\quad\text{or}\quad \ln\left(1+\frac{1}{v_0+1}\right)

Potential wrong answers from Q15a)

i. Variation 1

\frac{1}{v(v-1)}\equiv\frac{a}{v}+\frac{b}{v-1}

1\equiv a(v-1)+bv

v=1\Rightarrow b=1

v=0\Rightarrow a=-1

\frac{1}{v(v-1)}\equiv\frac{1}{v-1}-\frac1v

\int_{v_0}^{v_0/2}\frac{1}{v(v-1)}\,dv=\int_0^s dx

\int_{v_0}^{v_0/2}\left(\frac1{v-1}-\frac1v\right)dv=\int_0^s dx

\left[\ln|v-1|-\ln|v|\right]_{v_0}^{v_0/2}=s

s=\left[\ln\left|\frac{v-1}{v}\right|\right]_{v_0}^{v_0/2}

=\ln\left|\frac{v_0-2}{v_0}\right|-\ln\left|\frac{v_0-1}{v_0}\right|

=\ln\left|\frac{v_0-2}{v_0-1}\right|

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Mechanics

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

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