A person of mass \(M=90 \mathrm{~kg}\) standing on a smooth horizontal plane of ice throws a body of mass…
A person of mass \(M=90 \mathrm{~kg}\) standing on a smooth horizontal plane of ice throws a body of mass \(m=10 \mathrm{~kg}\) horizontally on the same surface. If the distance between the person and body after \(10 \mathrm{~s}\) is \(10 \mathrm{~m}\), then the KE of the person (in \(\mathrm{J}\)) is
\(0.55 \mathrm{~J}\)
\(4.5 \mathrm{~J}\)
\(0.90 \mathrm{~J}\)
0
Solution
According to the question body travels \(10 \mathrm{~m}\) in \(10 \mathrm{~s}\), hence its velocity is given as
\(v=\frac{10 \mathrm{~m}}{10 \mathrm{~s}}=1 \mathrm{~m} / \mathrm{s}\)
\(\therefore\) According to conservation of linear momentum. Initial total moment \(=\) Final total moment
\(0+0=M \cdot v^{\prime}+m v\)
where, \(v^{\prime}\) is velocity of person.
\(\begin{aligned}
& \qquad v^{\prime}=-\frac{m v}{M}=-\frac{10 \times 1}{90}=-\frac{1}{9} \mathrm{~m} / \mathrm{s} \\
& \therefore \text { Kinetic energy of person }=\frac{1}{2} M v^2 \\
& =\frac{1}{2} \times 90 \times\left(-\frac{1}{9}\right)^2=0.55 \mathrm{~J}
\end{aligned}\)