Question

A train of mass $M$ is moving on a circular track of radius $R$ with constant speed $v$ . The length of train is half the perimeter of track. The linear momentum of the train will be

• A. $0$
• B. $2Mv/\pi$
• C. $MvR$
• D. $Mv$

Answer 1

Answer: (B)

The net momentum is along the $x-$ axis, as net momentum in $y-$ axis is zero.
Momentum of the element is,
$dp={\int }_{\theta =-90}^{+90}\left(dm\right)\left(vcos\theta \right)$
Elemental mass is given by, $dm=\frac{M}{\pi R}\times Rd\theta$
Total momentum is,
$\Rightarrow p={\int }_{-90^{\circ }}^{90^{\circ }}\left[\left(\frac{M}{\pi R}Rd\theta \right)\left(vcos\theta \right)\right]$
$\Rightarrow p=\left(\frac{Mv}{\pi }\right){\int }_{-90^{\circ }}^{90^{\circ }}\left(cos\theta d\theta \right)=\frac{2Mv}{\pi }$