Obtain Newton's second law for a system of particles and write its statement.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The total linear momentum of a system of particles is given by $\vec{p} = M\vec{v}_{cm}$,where $M$ is the total mass of the system and $\vec{v}_{cm}$ is the velocity of the centre of mass.
Taking the derivative with respect to time $t$ on both sides:
$\frac{d\vec{p}}{dt} = M \frac{d\vec{v}_{cm}}{dt}$
Since the acceleration of the centre of mass is $\vec{a}_{cm} = \frac{d\vec{v}_{cm}}{dt}$,we have:
$\frac{d\vec{p}}{dt} = M\vec{a}_{cm}$
According to the motion of the centre of mass,the net external force acting on the system is $\vec{F}_{ext} = M\vec{a}_{cm}$.
Therefore,substituting this into the equation,we get:
$\frac{d\vec{p}}{dt} = \vec{F}_{ext}$
This is Newton's second law for a system of particles.
Statement: The external force acting on a system of particles is equal to the rate of change of the total linear momentum of the system.

Explore More

Similar Questions

$A$ shell fired from a gun at an angle to the horizontal explodes in mid-air. Then the centre of mass of the shell fragments will move

Two particles of masses $m_1$ and $m_2$ initially at rest start moving towards each other under their mutual force of attraction. The speed of the centre of mass at any time $t$,when they are at a distance $r$ apart,is

Two persons of mass $m_1$ and $m_2$ are standing at the two ends $A$ and $B$ respectively,of a trolley of mass $M$ as shown. Choose the incorrect statement,if $m_1 = m_2 = m$ and both the persons jump one by one,then

The figure shows a man of mass $m$ standing at the end $A$ of a trolley of mass $M$ placed at rest on a smooth horizontal surface. The man starts moving towards the end $B$ with a velocity $u_{rel}$ with respect to the trolley. The length of the trolley is $L$. Choose the correct statement.

$A$ body of mass $m_1 = 4 \text{ kg}$ moves at $5 \hat{i} \text{ m/s}$ and another body of mass $m_2 = 2 \text{ kg}$ moves at $10 \hat{i} \text{ m/s}$. The kinetic energy of the centre of mass is:

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