Two masses $m_1$ and $m_2$ are connected by a massless spring of spring constant $k$ and unstretched length $l$. The masses are placed on a frictionless straight channel,which we consider our $X$-axis. They are initially at rest at $x=0$ and $x=l$,respectively. At $t=0$,a velocity of $v_0$ is suddenly imparted to the first particle. At a later time $t$,the centre of mass of the two masses is at

  • A
    $x=\frac{m_2 l}{m_1+m_2}$
  • B
    $x=\frac{m_1 l}{m_1+m_2}+\frac{m_1 v_0 t}{m_1+m_2}$
  • C
    $x=\frac{m_2 l}{m_1+m_2}+\frac{m_2 v_0 t}{m_1+m_2}$
  • D
    $x=\frac{m_2 l}{m_1+m_2}+\frac{m_1 v_0 t}{m_1+m_2}$

Explore More

Similar Questions

Discuss the motion of binary (double) stars in astronomy.

$A$ body of mass $2 \,kg$ is moving towards north with a velocity of $20 \,m \,s^{-1}$ and another body of mass $3 \,kg$ is moving towards east with a velocity of $10 \,m \,s^{-1}$. The magnitude of the velocity of the centre of mass of the system of the two bodies is

Two blocks of masses $2 \text{ kg}$ and $1 \text{ kg}$ respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in $2 \text{ s}$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

$A$ body falling vertically downwards under gravity breaks into two parts of unequal masses. The centre of mass of the two parts taken together

Explain translational motion by the given illustration.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo