JEE PHYSICS MCQ: SYSTEM OF PARTICLES

1 Medium

1 of 1

Application +4 | -1

Two blocks of masses 10 kg and 4 kg connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of

14\mathrm{\;m}{\mathrm{s}^{ - 1}}

to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

View answer and explanation

Correct option: C

10\mathrm{\;m}{\mathrm{s}^{ - 1}}

Option D (

10\mathrm{\;m}{\mathrm{s}^{ - 1}}

) is correct because it is the velocity of the center of mass calculated using the formula for the center of mass velocity.

More Questions from SYSTEM OF PARTICLES

Practice All Chapter MCQs →

It is easier for a swimmer jumping into water from a height to describe a loop in the air by A uniform thin rod

\mathrm{AB}

of length

\mathrm{L}

has linear mass density

\mu(x)=a+\frac{b x}{L}

, where

x

is measured from

\mathrm{A}

. If the

\mathrm{CM}

of the rod lies at a distance of

\left(\frac{7}{12}\right) \mathrm{L}

from

\mathrm{A}

, then a and

\mathrm{b}

1.0 \mathrm{~g}

is rotating in a circle of diameter

8 \mathrm{~cm}

. Its moment of inertia will be The angular velocity of a particle is given by

\omega = {t^4} + 2{t^3}

then its angular acceleration of the particle at

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)