Question

A motorcar of mass $1200\mathrm{kg}$ is moving along a straight line with a uniform velocity of $90{\mathrm{kmh}}^{-1}$. Its velocity is slowed down to $18{\mathrm{kmh}}^{-1}$ in $4\mathrm{s}$ by an unbalanced external force. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required.

Answer 1

Step 1: Given data:

Mass($\mathrm{m}$) of motorcar= $1200\mathrm{kg}$

Initial uniform velocity($\mathrm{u}$)= $90{\mathrm{kmh}}^{-1}$

Since, $1{\mathrm{kmh}}^{-1}=\frac{5}{18}{\mathrm{ms}}^{-1}$

Thus,

$$\begin{gathered} \mathrm{u}=90\times \frac{5}{18}{\mathrm{ms}}^{-1} \\ \mathrm{u}=5\times 5 \\ \mathrm{u}=25{\mathrm{ms}}^{-1} \end{gathered}$$

Final velocity($\mathrm{v}$)= $18{\mathrm{kmh}}^{-1}$

Thus, $\mathrm{v}$= $18\times \frac{5}{18}{\mathrm{ms}}^{-1}$

$\mathrm{v}$= $5{\mathrm{ms}}^{-1}$

Time($\mathrm{t}$)= $4\mathrm{s}$

Step 2: Calculation of acceleration($\mathrm{a}$):

Using the first equation of motion, $\mathrm{v}=\mathrm{u}+\mathrm{at}$ we get

$$\begin{gathered} 5{\mathrm{ms}}^{-1}=25{\mathrm{ms}}^{-1}+\mathrm{a}\times 4\mathrm{s} \\ 25+4\mathrm{a}=5 \\ 4\mathrm{a}=5-25 \\ \mathrm{a}=\frac{-20}{4} \\ \mathbf{}\mathbf{}\mathbf{a}\mathbf{=}\mathbf{-}\mathbf{5}{\mathbf{ms}}^{\mathbf{-}\mathbf{2}} \end{gathered}$$

(Negative sign indicates the retardation in the body.)

Step 3: Calculation of Change in momentum:

Momentum($\mathrm{p}$) = $\mathrm{Mass}\left(\mathrm{m}\right)\times \mathrm{velocity}\left(\mathrm{v}\right)\mathrm{of}\mathrm{the}\mathrm{body}$

Thus,

The initial momentum(${\mathrm{p}}_{\mathrm{i}}$) of the motorcar is given as;

$$\begin{gathered} {\mathrm{p}}_{\mathrm{i}}=1200\mathrm{kg}\times 25{\mathrm{ms}}^{-1} \\ {\mathbf{p}}_{\mathbf{i}}\mathbf{=}\mathbf{30000}\mathbf{kg}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

And the final momentum(${\mathrm{p}}_{\mathrm{f}}$) of the motorcar is given as;

$$\begin{gathered} {\mathrm{p}}_{\mathrm{f}}=1200\mathrm{kg}\times 5{\mathrm{ms}}^{-1} \\ {\mathbf{p}}_{\mathbf{f}}\mathbf{=}\mathbf{6000}\mathbf{kg}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

Thus, the change in momentum = ${\mathrm{p}}_{\mathrm{f}}-{\mathrm{p}}_{\mathrm{i}}$

Now,

$$\begin{gathered} {\mathrm{p}}_{\mathrm{f}}-{\mathrm{p}}_{\mathrm{i}}=6000\mathrm{kg}{\mathrm{ms}}^{-1}-30000\mathrm{kg}{\mathrm{ms}}^{-1} \\ {\mathbf{p}}_{\mathbf{f}}\mathbf{-}{\mathbf{p}}_{\mathbf{i}}\mathbf{=}\mathbf{-}\mathbf{24000}\mathbf{kg}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

Step 4: Calculation of force ($\mathrm{F}$):

According to the second law of motion;

$$\begin{gathered} \mathrm{F}=\frac{-24000\mathrm{kg}{\mathrm{ms}}^{-1}}{4\mathrm{s}} \\ \mathbf{F}\mathbf{=}\mathbf{-}\mathbf{6000}\mathbf{N} \end{gathered}$$$\mathrm{Force}\left(\mathrm{F}\right)=\frac{\mathrm{Change}\mathrm{in}\mathrm{momentum}({\mathrm{p}}_{\mathrm{f}}-{\mathrm{p}}_{\mathrm{i}})}{\mathrm{time}\left(\mathrm{t}\right)}$

Thus,

The acceleration= $\mathbf{a}\mathbf{=}\mathbf{-}\mathbf{5}{\mathbf{ms}}^{\mathbf{-}\mathbf{2}}$

Change in momentum= $\mathbf{-}\mathbf{24000}\mathbf{kg}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$

the magnitude of the force required= $\mathbf{-}\mathbf{6000}\mathbf{N}$