## A block of mass M with a semicircular track of radius R rest on a horizontal frictionless surface….

Q: A block of mass M with a semicircular track of radius R rest on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A. The cylinder slips in the semicircular frictionless track. How far the block moved when the cylinder reaches the bottom of the track?

Sol: Since no net horizontal force acts on the system, it conserves its horizontal momentum.

$\displaystyle \vec{P} = M \vec{v_b} + m \vec{v_c} = 0$ .

Because initially the system was stationary.

$\displaystyle M \vec{v_b} + m \vec{v_c} = 0$ , Where $\vec{v_b}$ & $\vec{v_c}$ are the horizontal component of velocities of the block and the cylinder respectively

vcb = velocity of the cylinder relative to the block = v (say)

$\displaystyle v_b = \frac{m v_{cb} sin\theta}{M + m}$ ( Numerically )

$\displaystyle \int_{0}^{t} v_b dt = \frac{m}{M+m}\int_{0}^{t} (v_{cb})_x dt$

where t = time taken by the cylinder to arrive at the bottom of the block& (vcb)x = horizontal component of velocity of the cylinder relative to the block.

$\displaystyle x_b = \frac{m x_{cb}}{M+m}$ ; putting xbc = (R-r) we obtain

$\displaystyle x_b = \frac{m}{M+m}(R-r)$ ; Where $\displaystyle \int_{0}^{t} (v_{cb})_x dt = (R-r)$ and $\displaystyle \int_{0}^{t} v_b dt = x_b$

## Two particles A and B of which lighter particle has mass m, are released from infinity….

Q: Two particles A and B of which lighter particle has mass m, are released from infinity. They move towards each other under their mutual force of attraction. If their speeds are v and 2 v respectively find the K.E. of the system.

Sol: Since the particles are moving under their mutual gravitation, net force acting on the system (m1 + m2) is equal to $\displaystyle \vec{F_1} + \vec{F_2} = 0$

Therefore the linear momentum of the system is constant

$\displaystyle | m_1\vec{v_1} + m_2 \vec{v_2} | = constant$

Since initially the particles were released from rest, Pi = 0

$\displaystyle P_f = m_1\vec{v_1} + m_2 \vec{v_2} = 0$

$\displaystyle m_2 = \frac{m_1 v_1}{v_2}$ (numerically)

$\displaystyle m_2 = \frac{m v}{2 v} = \frac{m}{2}$ ; where m1 = m , v1 = v & v2 = 2v

The KE of the system $K.E = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

$\displaystyle K.E = \frac{1}{2}m v^2 + \frac{1}{2}(m/2 ) (2v)^2$

$\displaystyle K.E = \frac{3}{2}m v^2$

## The center of mass of a solid hemisphere of radius 8 cm is x cm from the center of the flat surface . Then value of x is ….

q: The center of mass of a solid hemisphere of radius 8 cm is x cm from the center of the flat surface . Then value of x is ….

Click to See Solution :
Ans: (3)
Sol: The center of mass of a hemisphere from the flat surface from its center is 3R/8 , hence x = 3R/8

$\displaystyle x = \frac{3}{8} \times 8$

x = 3 cm

## The coordinate of center of mass of a uniform flag shaped lamina (thin flat plate ) of mass 4 kg ….

The coordinate of center of mass of a uniform flag shaped lamina (thin flat plate ) of mass 4 kg (the coordinate of same are shown in figure ) are

(a) (1.25 m , 1.50 m)

(b) (0.75 m , 0.75 m)

(c) (0.75 m , 1.75 m )

(d) (1 m , 1.75 m)

Click to See Solution :
Ans: (c)
Sol: Area of upper rectangle = 2 Ã— 1 = 2 m2

Area of lower rectangle = 1 Ã— 2 = 2 m2

From Symmetry , the co-ordinate of center of mass of the upper and lower rectangles are (1, 2.5) and (0.5 , 0) respectively . So x-coordinate of center of mass is

$\displaystyle x = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

$\displaystyle x = \frac{2 \times 1 + 2 \times 0.5}{2 + 2} = 0.75 m$

Y-coordinate of center of mass is

$\displaystyle x = \frac{2 \times 2.5 + 2 \times 1}{2 + 2}= 1.75 m$

## A man weighing 80 kg is standing in a trolley weighing 320 kg . The trolley is resting on a frictionless horizontal rails ….

Q: A man weighing 80 kg is standing in a trolley weighing 320 kg . The trolley is resting on a frictionless horizontal rails . If the man starts walking on the trolley with speed of 1 m/s , then after 4 sec his displacement relative to the ground will be

(a)5 m

(b) 4.8 m

(c) 3.2 m

(d) 3.0 m

Ans: (c)

Solution: On applying conservation of linear momentum ,

80 x 1 = (80 + 320)v

v = 0.2 m/s

Velocity of man with respect ot ground = 1.0 – 0.2 = 0.8 m/s

Displacement of man relative to the ground

= 0.8 Ã— 4 = 3.2 m

## Two bodies of different masses of 2 kg and 4 kg are moving with velocities 20 m/s and 10 m/s towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass

Q: Two bodies of different masses of 2 kg and 4 kg are moving with velocities 20 m/s and 10 m/s towards each other due to mutual gravitational attraction. What is the velocity of their center of mass

(a) 5 m/s

(b) 6 m/s

(c) 8 m/s

(d) Zero

Ans: (d)
Sol: $\large \vec{v_c} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2}}{m_1 + m_2}$

$\large v_c = \frac{2 \times 20 + 4 \times (-10)}{2 + 4}$

$\large v_c = \frac{40 – 40}{6} = 0$

## Two object of masses 200 g and 500 g possess velocities 10i m/s and ( 3i + 5j ) m/s respectively. The velocity of their centre of mass in m/s is

Q: Two object of masses 200 g and 500 g possess velocities 10 i m/s and ( 3i + 5j ) m/s respectively. The velocity of their centre of mass in m/s is

(a) 5i â€“ 25j

(b) 7/5 i â€“ 25j

(c) 5i + 25/7 j

(d) 25i â€“ 5/7 j

Ans: (c)
Sol: $\displaystyle \vec{v_{cm}} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2}}{m_1 + m_2}$

$\displaystyle \vec{v_{cm}} = \frac{200 \times 10 \hat{i} + 500 \times (3 \hat{i}+ 5 \hat{j})}{200 + 500}$

$\displaystyle \vec{v_{cm}} = \frac{20 \hat{i} + 5 (3 \hat{i}+ 5 \hat{j})}{7}$

$\displaystyle \vec{v_{cm}} = \frac{35 \hat{i} + 25 \hat{j}}{7}$

$\displaystyle \vec{v_{cm}} = 5 \hat{i} + \frac{25}{7}\hat{j}$

## A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass….

Q: A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be

(a) 20 m

(b) 9.9 m

(c) 10.1 m

(d) 10 m

Ans: (c)
Sol: In gravity free space net force = 0

On Applying conservation of momentum ,

50 x v = 0.5 x 2

v = 1/50 m/s

time taken by stone to reach the ground t = 10/2 = 5 sec

Distance travel by man in this time = (1/50)x5 = 0.1 m

The distance of the man above the floor will be = 10 + 0.1 = 10.1 m

## A 2 kg body and a 3 kg body are moving along the x-axis. At a particular instant the 2 kg body has a velocity of 3 m/s and ….

Q: A 2 kg body and a 3 kg body are moving along the x-axis. At a particular instant the 2 kg body has a velocity of 3 m/s and the 3 kg body has the velocity of 2 m/s. The velocity of the centre of mass at that instant is

(a) 5 m/s

(b) 1 m/s

(c) 0

(d) None of these

Ans: (d)
Sol: The velocity of center if mass is

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

$\displaystyle \vec{v_{cm}} = \frac{2 \times 3 + 3 \times 2}{2 + 3}$

$\displaystyle v_{cm} = \frac{6 + 6}{5} = \frac{12}{5} m/s$

## Three bricks each of length L and mass M are arranged as shown from the wall. The distance of the centre of mass of the system from the wall is

Q: Three bricks each of length L and mass M are arranged as shown from the wall. The distance of the centre of mass of the system from the wall is

(a) L/4

(b) L/2

(c) (3/2) L

(d) (11/12)L

Ans: (d)

Sol: Distance of COM of 1st , 2nd & third bricks (taking from bottom) from wall will be L/2 , L & 5L/4 respectively

Distance of COM of the System will be ,

$\displaystyle r_{com} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3}{m_1 + m_2 + m_3}$

$\displaystyle r_{com} = \frac{M \times \frac{L}{2} + M \times L + M \times \frac{5L}{4}}{M + M + M}$

$\displaystyle r_{com} = \frac{ \frac{L}{2} + L + \frac{5L}{4}}{3}$

$\displaystyle r_{com} = \frac{ \frac{3 L}{2} + \frac{5L}{4}}{3}$

$\displaystyle r_{com} = \frac{11 L}{12}$