A truck of mass $\mathrm{M}$ and a car of mass $\frac{\mathrm{M}}{10}$ moving with same momentum are brought…
A truck of mass $\mathrm{M}$ and a car of mass $\frac{\mathrm{M}}{10}$ moving with same momentum are brought to halt by the application of same breaking force. The ratio of the distances travelled by truck and car before they come to stop is
$1: 10$
$1: \sqrt{10}$
$100: 1$
$5: 1$
Solution
Mass of truck, $m_1=M$
Mass of car, $m_2=\frac{M}{10}$
Initial velocity of truck $=u_1$
Initial velocity of $c a r=u_2$
Momentum of car \& truck are same thus,
$
\begin{aligned}
& \mathrm{p}_{\mathrm{T}}=\mathrm{p}_{\mathrm{C}} \\
& \mathrm{m}_1 \mathrm{u}_1=\mathrm{m}_2 \mathrm{u}_2 \\
& \mathrm{M} \mathrm{u}_1=\left(\frac{\mathrm{M}}{10}\right) \mathrm{u}_2 \\
& \therefore \frac{\mathrm{u}_1}{\mathrm{u}_2}=\frac{1}{10}
\end{aligned}
$
Both car \& truck will come to rest, thus their final velocity is zero.
using $\mathrm{eq}^{\mathrm{n}}$ of motion, $\mathrm{v}^2-\mathrm{u}^2=2$ as
Rearranging, put $\mathrm{v}=0 \& \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}$
$
\mathrm{S}=-\frac{\mathrm{mu}^2}{2 \mathrm{~F}}
$
Ratio of distance travelled by truck \& car is given as
$
\begin{aligned}
& \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{\mathrm{m}_1 \mathrm{u}_1^2}{2 \mathrm{~F}} \times \frac{2 \mathrm{~F}}{\mathrm{~m}_2 \mathrm{u}_2^2}=\frac{\mathrm{p}_1(2 \mathrm{~F}) \mathrm{u}_1}{\left(\mathrm{p}_2\right)(2 \mathrm{~F}) \mathrm{u}_2} \\
& \text { since, } \mathrm{p}_1=\mathrm{p}_2 \\
& \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{\mathrm{u}_1}{\mathrm{u}_2}=\frac{1}{10}
\end{aligned}
$