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.

Numerical

$\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

jee

(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 × 1 = 2 m2

Area of lower rectangle = 1 × 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 . 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

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

Click to See Answer :
Ans: (c)
Sol: 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 × 4 = 3.2 m