Q: Four identical sphere each of radius 10 cm and mass 1 kg are placed on a horizontal surface touching one another so that their centres are located at the corners of square of side 20 cm. What is the distance of their centre of square of side 20 cm either sphere ?

−−Centre of mass lies at centre of squareCA=21202+202=102 cmLength of diagonal =BA=202+202=202

Q: A body initially at rest, breaks up into two pieces of masses 2M and 3M respectively, together having a total kinetic energy E. The piece of mass 2M, after breaking up, has a kinetic energy.

21m1v12+21m2v22=E212Mv12+213Mv22=E⇒Mv12+23Mu22=E…(i)From momentum conservation0=m1φ1+m2φ2=2M1+3M2⇒v1=−23v2…(ii)From eqn. (i) and (ii)M(−23v2)2+23Mv22=E49Mv22+23Mv22=E⇒415Mv22=E⇒25⋅23Mv22=E⇒E2=52E=Mv12+52E=E.E1=E−52E=53E

Q: Two blocks of masses 2 kg and 3 kg are connected by massless spring and are placed on smooth surface. If impulse is given to 2 kg block and block starts moving with velocity 10 m/s in (+ve) x direction, then velocity of 2 kg block w.r.t centre of mass is

Vcom=m1+m2m1v1+m2v2=2+32×10+3×0Vcom=520=+4m/sV1=V1−Vcom=10−4=+6 m/s(+x) direction

Q: A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is :-

−−3MV0=2MV2+MV13 V0=2 V2+V1120=2 V2+60⇒V2=30 m/sK.E.ΔK.E.=213MV0221MV12+212MV22−213MV02=3 V02V12+2 V22−3 V02=48003600+1800−4800=81

Q: Four particles of mass m1=2m, m2=4m,m3=m and m4 are placed at four corners of a square. What should be the value of m4 so that the centre of mass of all the four particles are exactly at the centre of the square? − −

Let side of square is a.xcm=2a=m1+m2+m3+m4m1×0+m2×a+m3×a+m4×asolve to get m4=3mycm=2a=m1+m2+m3+m4m1×0+m2×0+m3×a+m4×asolve to get m4=5mThey are contradicting, so it is not possible to have centre of mass at the centre of square.

Question 1

Which of the following statements are correct?(a) Centre of mass of a body always coincides with the centre of gravity of the body.(b) Centre of mass of a body is the point at which the total gravitational torque on the body is zero(c) A couple on a body produce both translational and rotational motion in a body.(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.

Accepted Answer

Centre of mass may or may not coincide with centre of gravity.

Question 2

Three particles of masses 1 kg, 2 kg and 3 kg are situated at the corners of an equilateral triangle of side b. The coordinates of the centre of mass are :− −

Accepted Answer

The co-ordinates of points A, B and C are (0,0,0),(b, 0,0) and (2b​,2b3​​,0) respectively. Now, as the triangle is in X Y-plane, i.e., z-co-ordinate of all the masses is zero, so ZCM​=0Now,​XCM​=1+2+31×0+2×b+3(b/2)​=127b​YCM​=1+2+31×0+2×0+33​(b/2)​=1233​b​​ So, the co-ordinates of the centre of mass are[127b​,1233​b​,0] − −

Question 3

Four particles of masses 1 kg,2 kg,3 kg and 4 kg are placed at the four vertices A,B,C and D of a square of side 1 m. The coordinates of centre of mass of the particles are :-−−

Accepted Answer

−−Assuming D as the origin. DC as x-axis and DA as y-axis we havem1​=1 kg,(x1​,y1​)=(0,1 m)m2​=2 kg,(x2​,y2​)=(1 m,1 m)m3​=3 kg,(x3​,y3​)=(1 m,0)m4​=4 kg,(x4​,y4​)=(0,0)Coordination of their CM are calculated as follows.xCM​= m1​+m2​+m3​+m4​m1​x1​+m2​x2​+m3​x3​+m4​x4​​=1+2+3+4(1)(0)+2(1)+3(1)+4(0)​105​=21​ m=0.5 mSimilarly, yCM​= m1​+m2​+m3​+m4​m1​y1​+m2​y2​+m3​y3​+m4​y4​​=1+2+3+4(1)(1)+2(1)+3(0)+4(0)​=103​ m=0.3 m∴(xCM​,yCM​)=(0.5 m,0.3 m)

Question 4

A rod of length 3 m and its mass per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at

Accepted Answer

−−Here ρ=kx where k is a constant mass of small element of dx length isdm=kx.dxxcm​=∫dm∫x⋅dm​=∫03​x⋅dx∫03​x(xdx)​=[2x2​]03​[3x3​]03​​=29​327​​=2

Question 5

An isolated particle of mass m is moving in a horizontal (x−y) plane along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses 4m​&43m​. An instant later, the smaller fragment is at y=+15 cm. The larger fragment at this instant is at -

Accepted Answer

Before explosion particle was moving along x-axis i.e. it has no y-component of velocity. Therefore the centre ofmass will not move in y-direction or we can say ycm​=0Now ycm​= m1​+m2​m1​y1​+m2​y2​​0=(4m​+43 m​)(4m​)(+15)+(43 m​)y​⇒y=−5 cm

Question 6

Four identical sphere each of radius 10 cm and mass 1 kg are placed on a horizontal surface touching one another so that their centres are located at the corners of square of side 20 cm. What is the distance of their centre of square of side 20 cm either sphere ?

Accepted Answer

−−Centre of mass lies at centre of squareCA=21​202+202​=102​ cmLength of diagonal =BA=202+202​=202​

Question 7

A body initially at rest, breaks up into two pieces of masses 2M and 3M respectively, together having a total kinetic energy E. The piece of mass 2M, after breaking up, has a kinetic energy.

Accepted Answer

21​m1​v12​+21​m2​v22​=E21​2Mv12​+21​3Mv22​=E⇒Mv12​+23​Mu22​=E…(i)From momentum conservation0=m1​φ1​+m2​φ2​=2M1​+3M2​⇒v1​=−23​v2​…(ii)From eqn. (i) and (ii)M(−23​v2​)2+23​Mv22​=E49​Mv22​+23​Mv22​=E⇒415​Mv22​=E⇒25​⋅23​Mv22​=E⇒E2​=52E​=Mv12​+52E​=E.E1​=E−52E​=53E​

Question 8

Two blocks of masses 2 kg and 3 kg are connected by massless spring and are placed on smooth surface. If impulse is given to 2 kg block and block starts moving with velocity 10 m/s in (+ve) x direction, then velocity of 2 kg block w.r.t centre of mass is

Accepted Answer

Vcom​=m1​+m2​m1​v1​+m2​v2​​=2+32×10+3×0​Vcom​=520​=+4m/sV1​=V1​−Vcom​=10−4=+6 m/s(+x) direction

Question 9

A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is :-

Accepted Answer

−−3MV0​=2MV2​+MV1​3 V0​=2 V2​+V1​120=2 V2​+60⇒V2​=30 m/sK.E.ΔK.E.​=21​3MV02​21​MV12​+21​2MV22​−21​3MV02​​=3 V02​V12​+2 V22​−3 V02​​=48003600+1800−4800​=81​

Question 10

Four particles of mass m1​=2m, m2​=4m,m3​=m and m4​ are placed at four corners of a square. What should be the value of m4​ so that the centre of mass of all the four particles are exactly at the centre of the square?  − −

Accepted Answer

Let side of square is a.xcm​=2a​=m1​+m2​+m3​+m4​m1​×0+m2​×a+m3​×a+m4​×a​solve to get m4​=3mycm​=2a​=m1​+m2​+m3​+m4​m1​×0+m2​×0+m3​×a+m4​×a​solve to get m4​=5mThey are contradicting, so it is not possible to have centre of mass at the centre of square.

Question 11

Which of the following points is the likely position of the centre of mass of the system shown in figure?− −

Accepted Answer

The position of centre of mass of the system shown in Fig. is likely to be at C. This is because lower part of the sphere containing sand is heavier than upper part of the sphere containing air.

Question 12

Consider a two particle system with particles having masses m1​ and m2​. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?
(a) m1​m2​​d
(b) m1​+m2​m1​​d
(c) m2​m1​​d
(d) d

Accepted Answer

Exp. c
To keep the centre of mass at the same position, velocity of centre of mass is zero, so
m1​+m2​m1​v1​+m2​v2​​=0
where, v1​ and v2​ are the velocities of particles 1 and 2 , respectively.
⇒m1​dtdr1​​+m2​dtdr2​​=0
[∵v1​=dtdr1​​ and v2​=dtdr2​​]
⇒m1​dr1​+m2​dr2​=0
of particles. dr2​ represent the change in displacement 
 Let  2nd particle has been displaced by distance x
⇒m1​(d)+m2​(x)=0 or x=−m2​m1​d​
Alternate Solution
xCM​=m1​+m2​m1​x1​+m2​x2​​
After moving m1​ through a distance d towards CM and to keep the position of CM unchanged, let d′ be the shift of m2​.
xCM​=m1​+m2​m1​(x1​−d)+m2​(x2​+d′)​
From Eqs. (i) and (ii), we get
m1​+m2​m1​x1​+m2​x2​​=m1​+m2​m1​(x1​−d′)+m2​(x2​+d)​
⇒−m1​d+m2​d′=0
∴d′=m2​m1​​d

Question 13

Consider a two particle system with particles having masses m1​ and m2​. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?
(a) m1​m2​​d
(b) m1​+m2​m1​​d
(c) m2​m1​​d
(d) d

Accepted Answer

Exp. (c)To keep the centre of mass at the same position, velocity of centre of mass is zero, som1​+m2​m1​v1​+m2​v2​​=0where, v1​ and v2​ are the velocities of particles 1 and 2 , respectively.⇒m1​dtdr1​​+m2​dtdr2​​=0[∵v1​=dtdr1​​ & v2​=dtdr2​​]⇒m1​dr1​+m2​dr2​=0 of particles. dr2​ represent the change in displacement  Let 2 nd particle has been displaced by distance x⇒m1​(d)+m2​(x)=0 or x=−m2​m1​d​Alternate SolutionxCM​=m1​+m2​m1​x1​+m2​x2​​    ..(i)After moving m1​ through a distance d towards CM and to keep the position of CM unchanged, let d′ be the shift of m2​.xCM​=m1​+m2​m1​(x1​−d)+m2​(x2​+d′)​From Eqs. (i) and (ii), we getm1​+m2​m1​x1​+m2​x2​​=m1​+m2​m1​(x1​−d′)+m2​(x2​+d)​⇒−m1​d+m2​d′=0∴d′=m2​m1​​d

Question 14

A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mass 31​M and a body C of mass 32​m. The centre of mass of bodies B and C taken together shifts compared to that of body A towards(a) depends on height of breaking(b) does not shift(c) body C(d) body B

Accepted Answer

Exp. (b)Since, the acceleration of centre of mass in both the cases is same equal to g, so the centre of mass of the bodies B and C taken together does not shift compared to that of body A.

Question 15

A uniform slender rod 1 m long is initially standing vertically on a smooth horizontal surface. It is struck by a sharp horizontal blow at the top end, with the blow directed at right angles to the rod axis. As a result, the rod acquires an angular velocity of 3.00rad/s. What is the translational velocity of the centre of mass of the rod after the blow?(a) 0.25 m/s(b) 0.64 m/s(c) 0.5 m/s(d) 1.2 m/s

Accepted Answer

Let the magnitude of the impulse delivered by the blow be represented by p. From the translational and rotational forms of Newton's second law,vc​=Mp​ and Ic​ωc​=2pl​Here, vc​ is the rightward velocity of the centre of mass after the blow and ωc​ is the (positive clockwise) angular velocity of the rod after the blow about the point C.Using the fact that Ic​=12MI2​, we find thatωc​I=M6p​Therefore,vc​=6(ωc​/)​Since, ωc​=3.00rad/s and I=1.00 m, we findvc​=0.5 m/s

Question 16

Four particles A,B,C and D with masses mA​=m,mB​=2m,mC​=3m and mD​=4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles(in ms−2 ) is (a) 5a​(i^−j^​)(b)a(i^+j^​)(c) zero(d) 5a​(i^−j^​)(b)a(i^+j^​)

Accepted Answer

Exp. (a)For a system of discrete masses, acceleration of centre of mass (CM) is given byaCM​=mA​+mB​+mC​+mD​mA​aA​+mB​aB​+mC​aC​+mD​aD​​where, mA​=m1​mB​=2m1​mC​=3m and mD​=4m, ∣aA​∣=∣aB​∣=∣aC​∣=∣aD​∣=a (according to the question)aCM​=m+2m+3m+4m−mai^+2mai^+3mai^−4maj^​​=102ai^−2aj^​​=5a​⋅(i^−j^​)ms−2

Question 17

- A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be (a) (32a​,32b​)(b) (125a​,125b​)(c) (43a​,43b​)(d) (35a​,35b​)

Accepted Answer

Exp. (b)The given rectangular thin sheet ABCD can be drawn as shown in the figure below,Here,Area of complete lamina, A1​=abArea of shaded part of lamina =2a​×2b​=4ab​(x1​,y1​)= coordinates of centre of mass ofcomplete lamina =(2a​,2a​)(x2​,y2​)= coordinates of centre of mass ofshaded part of lamina =(43a​,43a​)∴ Using formula for centre of mass, we haveXCM​=A1​−A2​A1​X1​−A2​X2​​=ab−4ab​ab(2a​)−4ab​(43a​)​=43ab​168a2b−3a2b​​=125a​Similarly, YCM​=A1​−A2​A1​y1​−A2​y2​​=ab−4ab​ab(2b​)−4ab​(43b​)​=125b​Alternate SolutionLet m be the mass of entire rectangular lamina.So, the mass of the shaded portion of lamina=4m​Using the relation,XCM​=m1​−m2​m1​x1​−m2​x2​​, we getXCM​=m−4m​m(2a​)−4m​(43a​)​=43​2a​−163a​​=125a​Similarly, YCM​=m1​−m2​m1​y1​−m2​y2​​, we getYCM​=m−4m​m(2a​)−4m​(43b​)​=43​2b​−163b​​=125b​∴ The coordinates of the centre of mass of remaining portion will be (125a​,125b​).

Question 18

Three particles of masses 50 g,100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The (x,y) coordinates of the centre of mass will be (a) (43​​ m,125​ m)(b) (127​ m,83​​ m)(c) (127​ m,43​​ m)(d) (83​​ m,127​ m)

Accepted Answer

Exp⋅(c)The height of equilateral Δ ish=y3​=(1)2−(0.5)2​=3​/2 mThus, coordinates of three masses are (0,0),(1,0) and (0.5,23​​)Using,XCM​=m1​+m2​+m3​m1​x1​+m2​x2​+m3​x3​​,we haveXCM​=50+100+15050×0+100×1+150×0.5​=300175​=127​ mSimilarly,YCM​=m1​+m2​+m3​m1​y1​+m2​y2​+m3​y3​​=50+100+15050×0+100×0+150×23​​​=43​​ m

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