The given situation can be shown as :
Distance between H and Cl atoms = 1.27 \(\AA\)
Mass of H atom = \(m\)
Mass of Cl atom = \(35.5\, m\)
Let the centre of mass of the system lie at a distance x from the CI atom
Distance of the centre of mass from the H atom = (\(1.27-x\))
Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
\(\frac{m(1.27-x)+35.5mx}{m+35.5m}\) = \(0\)
\(m(1.27-x)+35.5\,mx\) = \(0\)
\(1.27-x\) = \(35.5x\)
\(\therefore\) \(x\) = \(\frac{-1.27}{(35.5-1)}\) = \(-0.037 \AA\)
Here, the negative sign indicates that the centre of mass lies at the left of the molecule.
Hence, the centre of mass of the HCl molecule lies \(0.037 \AA\) from the Cl atom.
Find the value of m if \(M = 10\) \(kg\). All the surfaces are rough.
As per the given figure, a small ball $P$ slides down the quadrant of a circle and hits the other ball $Q$ of equal mass which is initially at rest Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball $Q$ after collision will be :$\left( g =10 \,m / s ^2\right)$
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.
The formula for the Centre of Mass:
The imaginary point through which on an object or a system, the force of Gravity is acted upon is known as the Centre of Gravity of that system. Usually, it is assumed while doing mechanical problems that the gravitational field is uniform which means that the Centre of Gravity and the Centre of Mass is at the same position.