**1** **Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?**

Geometric centre; No$\\$ The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres. The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.

**2** **the HCl molecule, the separation between the nuclei of the two atoms is about $1.27 Å (1 Å = 10 -10 m )$ . Find the approximate location of the CM of the molecule, given that a chlorine atom is about $35.5$ times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.**

The given situation can be shown as:$\\$ Distance between H and Cl atoms =$1.27 Å$ $\\$ 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 Cl atom.$\\$ Distance of the centre of mass from the H atom $=(1.27 -x )$$\\$ Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:$\\$ $\dfrac{m(1.27-x)+35.5 mx}{m+35.5 m}=0\\ m(1.27-x)+35.5 mx=0\\ 1.27-x=-35.5x\\ \therefore x=\dfrac{-1.24}{35.5-1}=-0.037 A$$\\$ 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 $0.037Å $ from the Cl atom.

**3** **A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?**

No change$\\$ The child is running arbitrarily on a trolley moving with velocity v. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the child’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the (child+trolley) system, the child’s motion will produce no change in the speed of the centre of mass of the trolley

**4** **Show that the area of the triangle contained between the vectors $a$ and $b$ is one half of the magnitude of $a * b$ .**

Consider two vectors $\overrightarrow{OK}=|\vec{a}|$ and $\overrightarrow{OM}=|\vec{b}|,$ inclined at an angle $\theta $ , as shown in the following figure.$\\$ In $\Delta OMN,$ we can write the relation:$\\$ $\sin \theta =\dfrac{MN}{OM}=\dfrac{MN}{|\vec{a}|}\\ \therefore MN=|\vec{b}|\sin \theta \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\vec{a}*\vec{b}|=|\vec{a}||\vec{b}|\sin \theta\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\vec{a}*\vec{b}|=OK.MN*\dfrac{2}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\vec{a}*\vec{b}|=2*\text{Area of $\Delta OMK$}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \text{Area os } |\Delta OMK=\dfrac{1}{2}|\vec{a}*\vec{v}|$

**5** **Show that $a. (b × c)$ is equal in magnitude to the volume of the parallel piped formed on the three vectors, $a, b$ and $c.$**

A parallel piped with origin $O$ and sides $a, b$, and $c$ is shown in the following figure.$\\$ Volume of the given parallel piped = $abc$$\\$ $\overrightarrow{OC}=\vec{a}\\ \overrightarrow{OB}=\vec{b}\\ \overrightarrow{OC}=\vec{c}$$\\$ Let $n$ be a unit vector perpendicular to both $b$ and $c$. Hence, $n$ and $a$ have the same direction.$\\$ $\therefore \vec{b}*\vec{c}=bc \sin \theta n\\ =bc \sin 90^o n\\ =bcn\\ \vec{a}.(\vec{b}*\vec{c})=a.(bcn)\\ =abc \cos \theta n\\ =abc \cos 0^o\\ =abc\\ =\text{Volume of the parallelepiped}$