Q93 / 125 · BITSAT 2013 · Physics
A $3.628 \mathrm{~kg}$ freight car moving along a horizontal rail road spur track at $7.2 \mathrm{~km} /$ hour strikes a bumper whose coil springs experiences a maximum compression of $30 \mathrm{~cm}$ in stopping the car. The elastic potential energy of the springs at the instant when they are compressed $15 \mathrm{~cm}$ is
- A$12.1 \times 10^{4} \mathrm{~J}$
- B$121 \times 10^{4} \mathrm{~J}$
- Correct answerC$1.21 \times 10^{4} \mathrm{~J}$
- D$1.21 \times 10^{6} \mathrm{~J}$
Solution
\(\begin{aligned} 7.2 \mathrm{~km} / \mathrm{ram} &=\frac{7.2 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}}{7 / 2^{2}} \\ &=\frac{72 \times 5}{10 \times 18}=2 \mathrm{~m} / \mathrm{s} \end{aligned}\)
firt case EC \(\frac{1}{2} k x^{2}=\frac{1}{2} m v^{2}\) \(\Rightarrow \frac{1}{7} k\left(30 \times 10^{-2}\right)^{2}=\frac{1}{2} \times 3.628 \times(2)^{2}\) \(k=\frac{4 \times 3.628}{9 \times 10^{-2}}\)
\(\begin{aligned} \text { EC } & \frac{1}{2} k x^{2}=\frac{1}{2} m v^{2} \\ \Rightarrow & \frac{1}{2} k\left(30 \times 10^{-2}\right)^{2}=\frac{1}{2} \times 3.628 \times(2)^{2} \\ k &=\frac{4 \times 3.628}{9 \times 10^{-2}} \end{aligned}\)
energy potential enrgy when conpressed at \(15 \mathrm{~cm}\)
\(\begin{aligned}\Rightarrow & \frac{1}{2} k\left(15 \times 10^{-2}\right)^{2} \\=& \frac{1}{2} \times \frac{4 \times 3.628}{9 \times 10^{-2}} \times 225 \times 10^{-4} \\=& \frac{1639 \times 10^{-2}}{9} \\=& 182.206 \times 10^{-2} \end{aligned}\)
\(\begin{aligned} &\Rightarrow \frac{1}{2} k\left(15 \times 10^{2}\right. \\ &=\frac{1}{2} \times \frac{\alpha \times 3.628}{9 \times 10^{-2}} \times 225 \times 10^{-4} \\ &=\frac{1639 \times 10^{-2}}{9} \\ &=182.206 \times 10^{-2} \end{aligned}\)
Hence, the correct answer is option: 'C'.