Two persons of masses $55\, kg$ and $65\, kg$ respectively,are at the opposite ends of a boat. The length of the boat is $3.0\, m$ and it weighs $100\, kg$. The $55\, kg$ man walks up to the $65\, kg$ man and sits with him. If the boat is in still water,the center of mass of the system shifts by ....... $m$.

The equation $M\vec{A} = \vec{F}_{ext}$ denotes what?

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

$A$ 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:

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

Show that the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the centre of mass.