Search any question & find its solution
Question:
Answered & Verified by Expert
A train is moving along the tracks at a constant speed \(\mathrm{u}\). A girl on the train throws a ball of mass \(m\) straight ahead along the direction of motion of the train with speed \(v\) with respect to herself. Then
Options:
Solution:
2037 Upvotes
Verified Answer
The correct answers are:
Kinetic energy of the ball as measured by the girl on the train is \(\mathrm{mv}^2 / 2\), Work done by the girl in throwing the ball is \(\mathrm{mv}^2 / 2\), Work done by the train is mvu
Hint : w.r.t. the girl \(E_k=\frac{1}{2} m v^2\)
\(\therefore \mathrm{W}=\Delta \mathrm{E}_{\mathrm{k}}=1 / 2 \mathrm{mv}^2\)
Work by the train \(=\left\{\frac{1}{2}(v+u)^2-\frac{1}{2} m u^2\right\}-\frac{1}{2} m v^2\)
\(\begin{aligned}
& =\frac{1}{2} m\left(v^2+u^2+2 v u\right)-\frac{1}{2} m\left(v^2+u^2\right) \\
& =m v u
\end{aligned}\)
Gain in \(E_k=\frac{1}{2} m(v+u)^2-\frac{1}{2} m u^2=\frac{1}{2} m v^2+m v u\)
measured from rail track
\(\therefore \mathrm{W}=\Delta \mathrm{E}_{\mathrm{k}}=1 / 2 \mathrm{mv}^2\)
Work by the train \(=\left\{\frac{1}{2}(v+u)^2-\frac{1}{2} m u^2\right\}-\frac{1}{2} m v^2\)
\(\begin{aligned}
& =\frac{1}{2} m\left(v^2+u^2+2 v u\right)-\frac{1}{2} m\left(v^2+u^2\right) \\
& =m v u
\end{aligned}\)
Gain in \(E_k=\frac{1}{2} m(v+u)^2-\frac{1}{2} m u^2=\frac{1}{2} m v^2+m v u\)
measured from rail track
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.