## A 10 kg boy standing in a 40 kg boat floating on water is 20 m from the shore of the river. If he moves 8 metres on the boat…

Q: A 10 kg boy standing in a 40 kg boat floating on water is 20 m from the shore of the river. If he moves 8 metres on the boat towards the shore, then how far is he from the shore now ?

Sol: Mass of the body (m) = 10 kg Mass of the boat (M) = 40 kg

Distance travelled by boy (l) = 8 m

distance travelled by the boat in opposite direction

$\large = \frac{m l}{M + m}$

$\large = \frac{10 \times 8}{40 + 10}$

= 1.6 m

Distance of the body from the shore is (20 – 8) + (1.6) = 13.6m

## Two particles A and B initially at rest, move towards each other, under mutual force of attraction…

Q: Two particles A and B initially at rest, move towards each other, under mutual force of attraction. At an instance when the speed of A is v and speed of B is 2v, the speed of centre of mass (CM) is

Sol: Internal forces doesn’t change the position of centre of mass. So velocity of CM is zero.

## Two bodies of 6kg and 4 kg masses have their velocity $5\hat{i} – 2\hat{j} + 10 \hat{k}$ and $10\hat{i} – 2\hat{j} + 5 \hat{k}$ respectively.

Q: Two bodies of 6kg and 4 kg masses have their velocity $5\hat{i} – 2\hat{j} + 10 \hat{k}$ and $10\hat{i} – 2\hat{j} + 5 \hat{k}$ respectively. Then, the velocity of their centre of mass is

Sol: The velocity of centre of mass is

$\large \vec{v_{cm}} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2}}{m_1 + m_2}$

$\large \vec{v_{cm}} = \frac{6 ((5\hat{i} – 2\hat{j} + 10 \hat{k})) + 4 (10\hat{i} – 2\hat{j} + 5 \hat{k})}{6 + 4}$

$\large = 7 \hat{i} – 2\hat{j} + 8\hat{k}$

## An object A is dropped from the top of a 30 m high building and at the same moment another object B is projected vertically upwards…

Q:  An object A is dropped from the top of a 30 m high building and at the same moment another object B is projected vertically upwards with an initial speed of 15 m/s from the base of the building. Mass of the object A is 2kg while mass of the object B is 4 kg. Find the maximum height about the ground level attained by the centre of mass of A and B system (take g = 10 m/s2)

Sol: m1 = 4 kg, m2 = 2 kg. Initially 4 kg is on the ground, therefore x1 = 0 and 2 kg is on top of the building,
therefore x2 = 30m.

$\large x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

$\large = \frac{0 + 2 \times 30}{4 + 2} = 10 m$

Initial velocity of CM,

$\large u_{cm} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

$\large u_{cm} = \frac{4 \times 15 + 0}{4+2}$

=10 m/s upward

acceleration of CM,

aCM = g = 10m/s2 downwards.

Maximum height attained by CM from initial position,

$\large H_{cm} = \frac{u_{cm}^2}{2 g} = \frac{10^2}{20} = 5 m$

## If the centre of mass of three particles of masses, of 1 kg, 2 kg, 3kg is at (2,2,2), then where should a fourth particle of mass…

Q: If the centre of mass of three particles of masses, of 1 kg, 2 kg, 3kg is at (2,2,2), then where should a fourth particle of mass 4kg be placed so that the combined centre of mass may be at (0, 0, 0).

Sol: Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the positions of masses 1 kg, 2 kg, 3 kg and let the co-ordinates of centre of mass of the three particle system is (xcm,ycm, zcm) respectively.

$\large x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$

$\large 2 = \frac{1 \times x_1 + 2 \times x_2 + 3 \times x_3}{1 + 2 + 3}$

$\large x_1 + 2 x_2 + 3 x_3 = 12$ …(i)

Suppose the fourth particle of mass 4kg is placed at (x4, y4, z4) so that centre of mass of new system shifts to (0, 0, 0). For x coordinate of new centre of mass we have

$\large 0 = \frac{1 \times x_1 + 2 \times x_2 + 3 \times x_3 + 4 \times x_4 }{1 + 2 + 3 + 4}$

$\large x_1 + 2 x_2 + 3 x_3 + 4 x_4 = 0$ …(ii)

From equations (i) and (ii)

12 + 4x4 = 0 ⇒ x4 = -3

Similarly, y4 = -3 and z4 = -3

Therefore 4kg should be placed at (-3, -3, -3).

## When ‘n’ number of particles each of mass ‘m’ are at distances x1 = a, x2 = ar, x3 = ar^2 , ……. xn = ar^n , units from origin…

Q: When ‘n’ number of particles each of mass ‘m’ are at distances x1 = a, x2 = ar, x3 = ar2 , ……. xn = arn , units from origin one the X – axis, then find the distance of their centre of mass from origin.

Sol: $\large x_{cm} = \frac{m \times a + m \times ar + …. + m \times ar^n}{m + m + m + ….+ n terms}$

$\large = \frac{m(a+ ar + ar^2 + …+ ar^n)}{n m}$

If r >1 ;

$\large x_{cm} = \frac{1}{n} [\frac{a(r^n -1)}{r-1}]$

If x < 1 $\large x_{cm} = \frac{1}{n} [\frac{a(1- r^n)}{1-r}]$

## When ‘n’ number of particles of masses m, 2m, 3m, …..nm are at distances x1 = 1, x2 = 4, x3 = 9, ….xn = n2 units respectively…

Q: When ‘n’ number of particles of masses m , 2m , 3m , …..nm are at distances x1 = 1, x2 = 4, x3 = 9, ….xn = n2 units respectively from origin on the X – axis, then find the distance of their centre of mass from origin.

Sol: $\large x_{cm} = \frac{m \times 1 + 2m \times 4 + 3m \times 9 + ….+ nm \times n^2}{m + 2m + 3m + …+mn}$

$\large x_{cm} = \frac{1^3 + 2^3 + 3^3 + ….+ n^3}{1 + 2 + 3 +….+ n}$

$\large = \frac{(\frac{n(n+1)}{2})^2}{\frac{n(n+1)}{2}}$

$\large = \frac{n(n+1)}{2}$