Question

A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of $60cm{s}^{-1}$on the string when the car is at rest and $62cm{s}^{-1}$ when the car accelerates on a horizontal road. Find the acceleration of the car. Take $g=10m{s}^{-2}$.

Answer 1

Step 1: Given Data:

Speed at rest, $v_1=60cm{s}^{-1}$

Speed when car is in motion, $v_2=62cm{s}^{-1}$

Step 2: Formulas used:

When car is at rest,

Tension, $T_1=mg$

Speed, $v_1=\sqrt{T_1/\rho }=\sqrt{(mg/\rho )}$……….….$\left(1\right)$

When car is in motion,

Tension, $T_2=m\sqrt{(a^2+g^2)}$

Speed, $v_2=\sqrt{T_2/\rho }=\sqrt{\left[\right(m\sqrt{(a^2+g^2)})/\rho ]}$……….….$\left(2\right)$

Step 3: Finding the acceleration of the car:

Now,

$\frac{v_2}{v_1}=\frac{\sqrt{{\displaystyle \frac{m\sqrt{a^2+g^2}}{\rho }}}}{\sqrt{{\displaystyle \frac{mg}{\rho }}}}$

$$\begin{gathered} \Rightarrow \\ \Rightarrow \\ \Rightarrow \\ \Rightarrow \\ \Rightarrow \\ \Rightarrow \\ \Rightarrow \end{gathered}$$$\begin{array}{rcl}{\left(\frac{v_2}{v_1}\right)}^4& =& \frac{a^2+g^2}{g^2}\\ {\left(\frac{62}{60}\right)}^4& =& \frac{a^2+g^2}{g^2}\\ {\left(1.03\right)}^4& =& \frac{a^2+g^2}{g^2}\\ 1.140x{g}^2& =& a^2+g^2\\ g^2\left(1.140-1\right)& =& a^2\\ 10x\sqrt{0.140}& =& a\\ a& =& 3.7m/s\end{array}$

Hence, the acceleration of the car is $3.7m/s$