Question.
A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Solution:
(i) Initial velocity of the car, $\mathrm{u}=90 \mathrm{~km} / \mathrm{h}=25
\mathrm{~ms}^{-1}$; final velocity of the car, $\mathrm{v}=18 \mathrm{~km} / \mathrm{h}=$
$5 \mathrm{~ms}^{-1} ;$ time, $\mathrm{t}=4 \mathrm{~s} ;$ acceleration, $\mathrm{a}=?$
We know
$\mathrm{v}=\mathrm{u}+\mathrm{at}$
$5=25+a \times 4$
$\therefore \quad-\mathrm{a} \times 4=20$ or $\mathrm{a}=\frac{-20}{4}=-5 \mathrm{~ms}^{-2}$
(ii) Change in momentum, $\Delta p=m(v-u)$
$=1200(5-25)=1200 \times(-20)=-24000 \mathrm{Ns}$
(iii) Magnitude of force
$\mathrm{F}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}=\frac{-24000}{4}=-6000 \mathrm{~N}$
(i) Initial velocity of the car, $\mathrm{u}=90 \mathrm{~km} / \mathrm{h}=25
\mathrm{~ms}^{-1}$; final velocity of the car, $\mathrm{v}=18 \mathrm{~km} / \mathrm{h}=$
$5 \mathrm{~ms}^{-1} ;$ time, $\mathrm{t}=4 \mathrm{~s} ;$ acceleration, $\mathrm{a}=?$
We know
$\mathrm{v}=\mathrm{u}+\mathrm{at}$
$5=25+a \times 4$
$\therefore \quad-\mathrm{a} \times 4=20$ or $\mathrm{a}=\frac{-20}{4}=-5 \mathrm{~ms}^{-2}$
(ii) Change in momentum, $\Delta p=m(v-u)$
$=1200(5-25)=1200 \times(-20)=-24000 \mathrm{Ns}$
(iii) Magnitude of force
$\mathrm{F}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}=\frac{-24000}{4}=-6000 \mathrm{~N}$
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