KKMishra's Tutorials: Solutions to Problems on "CENTER OF MASS, LINEAR MOMENTUM, COLLISION" - H C Verma's Concepts of Physics, Part-I, Chapter-9, EXERCISES-Q1-Q10

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CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
EXERCISES Q-1 to Q-10

1. Three particles of masses 1.0 kg, 2.0 kg and 3.0 kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1 m. Locate the center of mass of the system.

Answer: Let us draw a diagram as below,

Diagram for Q-1

We take point B as origin and BC as the x-axis.

m = 1.0 kg. m' = 2.0 kg m" = 3.0 kg. Total mass M = 6.0 kg.

If the coordinate of CoM be (X, Y) then

X = (1/M)*(1.0 kg*0.5 m+2.0 kg*0 m+3.0 kg*1.0 m)

=(1/6)*(0.5+0+3) = 3.5/6 = 35/60 = 7/12 m

Y = (1/M)*(1.0 kg*1.0*sin60° m+2.0 kg*0+3.0 kg*0)

= (1/6)*(√3/2) = √3/12 m

Hence assuming B as origin and BC as the x-axis the coordinate of CoM will be (7/12 m, √3/12 m).

2. The structure of a water molecule is shown in figure (9-E1). Find the distance of the center mass of the molecule from the center of the oxygen atom.

Answer: Let us take the mass of Hydrogen atom as 1 unit and Oxygen atom as 16 units. From the symmetry, it is clear that the CoM will lie on the perpendicular bisector of the line joining Hydrogen atoms which will pass through the oxygen atom. Let the distance of CoM from Oxygen atom be Y.

Figure for Q - 2

Taking a horizontal line passing through the center of Oxygen atom as an axis. OH = r, m = mass of hydrogen atom, m' = mass of Oxygen atom. M = 2+16 = 18 unit.

Y = (1/M)*(2*mr*cos52°+16*0)

= (1/18)*(2*1*0.96x10^-10*0.616 m)
= 0.066 x 10^-10m.
= 6.6 x 10^-12m
Hence distance of the center of mass of the molecule from the center of oxygen atom is 6.6 x 10^-12m

3. Seven homogeneous bricks, each of length L, are arranged as shown in figure (9-E2). Each brick is displaced with respect to the one in contact by L/10. Find the x-coordinate of the of the center of mass relative to the origin shown.

Answer: Let the mass of each brick be m units. Hence the total mass of the arrangement = 7m units. Since the bricks are homogeneous and rectangular, the CoM of each brick will be at its center and mass of each brick will be assumed to be concentrated at this CoM.
x - Coordinate of the center of 1st and 7th brick = L/2 units
x - Coordinate of the center of 2nd and 6th brick = L/2+L/10 =6L/10 = 3L/5 units
x - Coordinate of the center of 3rd and 5th brick = L/2+L/10+L/10
=7L/10 units
x - Coordinate of the center of 4th brick = L/2+L/10+L/10+L/10
=8L/10 =4L/5 units
Hence x - Coordinate of the center of mass of the arrangement, X
= (1/7m)*(2m*L/2+2m*3L/5+2m*7L/10+m*4L/5)
= (1+6/5+14/10+ 4/5)L/7
=(10+12+14+8)L/70
=44L/70
=22L/35 units.

4. A uniform disc of radius R is put over another uniform disc of radius 2R of the same thickness and density. The peripheries of the two discs touch each other. Locate the center of the mass of the system.

Answer: Since the discs are uniform and of same density and thickness, the masses of each circle will be proportional to its area and assumed to be concentrated at its center.

Mass of smaller circle m = πR²

Distance of CoM of smaller circle from the center of larger circle, x = R

Mass of Larger circle m' = π.(2R)² = 4πR²

Distance of CoM of larger circle from the center of larger circle, x' = 0

Total Mass, M = πR²+4πR² = 5πR²

Distance of center of mass of the system from the center of the larger circle, X = (1/5πR²)*(πR²*R+5πR²*0)

= R/5.

5. A disc of radius R is cut out from a larger disc of radius 2R in such a way that the edge of the hole touches the edge of the disc. Locate the center of mass of the residual disc.

Answer: The cut-out disc can be assumed as a negative mass,

m = -πR².

Its distance from the origin, x = R

Mass of the larger disc, m' = 4πR²

Its distance from origin, x' = 0.

Mass of the residual disc, M = 4πR²-πR² =3πR²

Figure for Q-5

Let the distance of the center of mass of the residual disc from the origin = X

X =(1/M)*(-πR²*R+4πR²*0)

=(1/3πR²)*(-πR³)

=-R/3

The negative sign says that the CoM of the residual disc is located away from the hole. And it is located at R/3 distance from the center of the larger disc.

6. A square plate of edge d and a circular disc of diameter d are placed touching each other at the midpoint of an edge of the plate as shown in figure (9-Q2). Locate the center of mass of the combination, assuming same mass per unit area for two plates.

Figure for Q - 6

Answer: Since the mass per unit area for both the plates are same, their masses will be proportional to their areas. Masses of each object will be assumed to be concentrated at their geometrical centers which are also their CoM. Let the center of the disc be the origin.
Mass of disc, m = πd²/4, Distance from the origin = 0
Mass of the square plate, m' = d², Distance from the origin = d.
Total mass of the system, M = πd²/4+d²
Let the distance of the center of mass of the system from the origin be X.
X = {1/(πd²/4 +d²)}*(πd²/4*0+d²*d)
= d³/(πd²/4+d²)
= d/(π/4+1)
=4d/(π+4)
So the center of mass of the system will be 4d/(π+4) unit distance right to the center of the disc.

7. Calculate the velocity of the center of mass of the system of particles shown in figure (9-E3).

Figure for Q-7

Answer: Let the mass, velocity and the position of the particles with respect to the origin be as follows,

m₁ = 1.0 kg, at distance r₁ v₁ = -1.5*cos37°i-1.5*sin37°j
m₂ = 1.2 kg, at distance r₂ v₂ = 0.4 j
m₃ = 1.5 kg, at distance r₃ v₃ = -1.0*cos37°i+1.0*sin37°j
m₄ = 0.5 kg, at distance r₄ v₄ = 3.0 i
m₅ = 1.0 kg, at distance r₅ v₅ = 2.0*cos37°i-2.0*sin37°j

Total mass of the system, M = 1.0+1.2+1.5+0.5+1.0 =5.2 kg

Position of the center of mass of the system, X
= (1/M)*Σmr = (1/M)*Σm(xi+yj)
Hence the velocity of the center of mass,
V = dX/dt = (1/M)*Σm(v_xi+v_yj)
=(1/5.2)*[1.0*(-1.5*cos37°i-1.5*sin37°j)+1.2*(0.4j) +1.5*(-1.0*cos37°i+1.0*sin37°j) +0.5*(3.0i) +1.0*(2.0*cos37°i-2.0*sin37°j)]
=(1/5.2)*[(-1.5*cos37°-1.5*cos37°+1.5+2.0*cos37°)i +(-1.5*sin37°+0.48+1.5*sin37°-2sin37°)j]
=(1/5.2)*[(1.5-cos37°)i+(0.48-2sin37°)j]
=(1/5.2)*(0.7 i - 0.72 j)
= 0.134 i - 0.14 j
V = √(0.134²+0.14²) =√0.038 = 0.194 m/s ≈ 0.20 m/s
Tanθ = -0.14/0.134 = -1.04
θ ≈ -46°
Hence velocity of the CoM is 0.20 m/s at an angle 46° down towards right.

8. Two blocks of masses 10 kg and 20 kg are placed on the x-axis. The first mass is moved on the axis by a distance of 2 cm. By what distance should the second mass be moved to keep the position of the center of mass unchanged.

Answer: Let the origin be at 10 kg block. The distance of the CoM from 10 kg block be x cm and the distance of 20 kg block from origin be y cm.

x = (1/30) (10*0 + 20*y) =2y/3

When the 10 kg block is moved on the axis by a distance of 2 cm and the CoM remains unchanged then let the position of 20 kg block be y'.

x = (1/30) (10*2+20*y')

→20+20y' = 30 x

→2y' =3x - 2

→y' = 1.5x-1 = 1.5*2y/3 -1 =y-1 cm

So, to keep the CoM unchanged, the 20 kg block should be moved by 1 cm in the opposite direction.

9. Two blocks of masses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the center of mass by 1 cm.

Answer: Let the 30 kg block be at the origin and the 10 kg block be at y cm above the origin.

The distance of the CoM from the origin, x = (1/40)*{30*0+10*y}

=y/4 cm.

When the 10 kg block is raised through a height of 7 cm, new distance from the origin is y+7 cm. Let the distance of 30 kg block from the origin after movement to keep the CoM 1 cm above the old position be x'.

Now x+1 =(1/40){30*x'+10(y+7)}

→y/4 + 1 =3x'/4 + y/4 + 7/4

→3x'/4 = 1-7/4 = -3/4

→x' = -1 cm

Hence the 30 kg block should be moved down by 1 cm to raise the CoM above by 1 cm.

10. Consider a gravity-free hall in which a tray of mass M, carrying a cubical block of ice of mass m and edge L, is at rest in the middle (Figure 9-E4). If the ice melts, by what distance does the center of mass of 'the tray plus the ice" system descend?

Figure for Q - 10

Answer: Since "the tray plus the ice" system is placed in a gravity-free hall, even when the ice melts there is no change in the shape of water (melted ice). Hence there will be no change in the position of the Center of the mass of "the tray and the ice" system. (Here we are not considering the effects of surface tension or adhesion).

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Links to the chapter -

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CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-15

EXERCISES - Q-16 TO Q-30

CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

Questions for Short Answers

Objective - I