Tuesday, August 24, 2021

Motion of systems of particles and rigid body: Physics

 



Q.1. The distance travelled by an object along a straight line in time t is given by s space equals space 8 t squared plus 3 t minus 5 The initial velocity of the object is 

A) 3 unit

B) -3 unit 

C) 5 unit 

D) -5 unit 

Explanation-

          s space equals space 8 t squared plus 3 t minus 5

         fraction numerator d s over denominator d t end fraction equals 16 space t space plus 3

Initial velocity, u=right enclose fraction numerator d s over denominator d t end fraction end enclose subscript t equals 0 end subscript equals 3 space u n i t space 


Q.2. Two particles of masses 1 kg and 3 kg have position vectors 2 straight i with hat on top plus 3 straight j with hat on top plus 4 stack straight k space with hat on top space and space minus 2 straight i with hat on top plus 3 straight j with hat on top minus 4 straight k with hat on top  respectively. The center of mass has a position vector 

A) straight i with hat on top plus 3 j with hat on top minus 2 k with hat on top

B) negative 1 with hat on top minus 3 straight j with hat on top minus 2 straight k with hat on top

C) negative straight i with hat on top plus 3 straight j with hat on top plus 3 straight k with hat on top

D) negative straight i with hat on top plus 3 straight j with hat on top minus 2 straight k with hat on top 

Explanation-

Here, straight m subscript 1 equals 1 space kg comma space straight m subscript 2 equals 3 space kg
stack straight r subscript 1 with rightwards arrow on top equals 2 straight i with hat on top plus 3 straight j with hat on top plus 4 straight k with hat on top comma space straight r with rightwards arrow on top subscript 2 equals negative 2 straight i with hat on top plus 3 straight j with hat on top minus 4 straight k with hat on top

The position vector of center of mass is 

straight R with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 r with rightwards arrow on top subscript 1 plus m subscript 2 r with rightwards arrow on top subscript 2 over denominator m subscript 1 plus m subscript 2 end fraction equals fraction numerator left parenthesis 1 right parenthesis left parenthesis 2 i with hat on top plus 3 j with hat on top plus 4 k with hat on top right parenthesis plus left parenthesis 3 right parenthesis space left parenthesis negative 2 i with hat on top plus 3 j with hat on top minus 4 k with hat on top right parenthesis over denominator 1 plus 3 end fraction
space space space space space space space space space space equals fraction numerator 2 i with hat on top plus 3 j with hat on top plus 4 k with hat on top minus 6 i with hat on top plus 9 j with hat on top minus 12 k with hat on top over denominator 4 end fraction equals fraction numerator negative 4 i with hat on top plus 12 j with hat on top minus 8 k with hat on top over denominator 4 end fraction
space space space space space space space space space space equals negative i with hat on top plus 3 j with hat on top minus 2 k with hat on top 


Q.3. Three bricks each of length L and M are arranged as shown from the wall. The distance of the center of mass of the system from the wall is

 

A) straight L over 4

B) straight L over 2

C) open parentheses 3 over 2 close parentheses L

D) open parentheses 11 over 12 close parentheses L 

Explanation-

From figure , 

straight x subscript 1 equals straight L over 2 semicolon space straight x subscript 2 equals straight L over 2 plus straight L over 2 equals straight L space semicolon space straight x subscript 3 equals straight L over 2 plus straight L over 4 plus straight L over 2 equals fraction numerator 5 straight L over denominator 4 end fraction
therefore space space space straight X subscript CM space end subscript equals space fraction numerator straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 plus straight m subscript 3 straight x subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction
space space space space space space space space space space space space space space space space space equals space fraction numerator straight M cross times begin display style straight L over 2 end style plus straight M cross times straight L plus straight M cross times begin display style fraction numerator 5 straight L over denominator 4 end fraction end style over denominator straight M plus straight M plus straight M end fraction equals space fraction numerator begin display style 11 over 4 end style ML over denominator 3 straight M end fraction equals 11 over 12 straight L 


Q.4. Three particles, each of mass m, are placed at the corners of a right angled triangle as shown in figure. If OA = a and OB = b , the position vector of the center of mass is (Here i with hat on top space a n d space j with hat on top   are unit vectors along x and y axes respectively ).

A) 1 third left parenthesis a i with hat on top space plus b j with hat on top right parenthesis space

B) 1 third space left parenthesis a i with hat on top space minus b j with hat on top

C) 2 over 3 space left parenthesis a i with hat on top space plus b j with hat on top right parenthesis

D) 2 over 3 space left parenthesis a i with hat on top space minus b j with hat on top right parenthesis 

Explanation-

The (x , y ) co- ordinates of the masses at O, A and B respectively are ( refer to figure as given in the question ) 

left parenthesis straight x subscript 1 space equals space 0 comma space straight y subscript 1 equals 0 right parenthesis comma space left parenthesis straight x subscript 2 equals straight a comma space straight y subscript 2 equals 0 right parenthesis space and space left parenthesis straight x subscript 3 equals 0 comma space straight y subscript 3 equals straight b right parenthesis space
The space left parenthesis straight x comma space straight y space right parenthesis space co minus ordinates space of space the space center space of space mass space are space
straight X subscript CM equals fraction numerator straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 plus straight m subscript 3 straight x subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction equals fraction numerator straight m cross times 0 plus straight m cross times straight a plus straight m cross times 0 over denominator straight m plus straight m plus straight m end fraction space equals straight a over 3
straight Y subscript CM space equals fraction numerator straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 plus straight m subscript 3 straight y subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction equals space fraction numerator straight m cross times 0 plus straight m cross times 0 plus straight m cross times straight b over denominator straight m plus straight m plus straight m end fraction equals space straight b over 3
The space position space vector space of space the space centre space of space mass space is space
straight R with rightwards arrow on top subscript CM equals straight X subscript CM straight i with hat on top plus straight Y subscript CM straight j with hat on top space equals space straight a over 3 straight i with hat on top plus straight b over 3 straight j with hat on top equals space 1 third left parenthesis straight a straight i with hat on top space plus straight b straight j with hat on top right parenthesis 


Q.5. In a carbon monoxide molecule , the carbon and oxygen atom are separated by a distance of 1.12 space cross times space 10 to the power of negative 10 end exponent m. The distance of the center of mass from the carbon atom is 

A) 0.48 space cross times 10 to the power of negative 10 end exponent straight m

B) 0.51 cross times 10 to the power of negative 10 end exponent straight m

C) 0.56 space cross times 10 to the power of negative 10 end exponent straight m

D) 0.64 space cross times 10 to the power of negative 10 end exponent straight m 

Explanation-

Since the mass of carbon atom is 12 units and that of oxygen atom is 16 units , the distance of the centre of mass from the carbon atom is 

straight R subscript CM equals fraction numerator 12 cross times 0 plus 16 cross times 1.12 cross times 10 to the power of negative 10 end exponent over denominator 12 plus 16 end fraction equals 0.64 cross times 10 to the power of negative 10 end exponent straight m 


Q.6. The centre of mass of three bodies each of mass 1 kg located at the points (0,0), (3,0) and (0, 4 ) in the x-y plane is 

A) open parentheses 4 over 3 comma 1 close parentheses

B) open parentheses 1 third comma 2 over 3 close parentheses

C) open parentheses 1 half comma 1 half close parentheses

D) open parentheses 1 comma 4 over 3 close parentheses 

Explanation-

Here , straight m subscript 1 equals 1 space kg comma space left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis space equals space left parenthesis 0 comma 0 right parenthesis comma space straight m subscript 2 equals space 1 kg comma space
left parenthesis straight x subscript 2 comma straight y subscript 2 right parenthesis space equals left parenthesis 3 comma 0 right parenthesis comma space straight m subscript 3 equals space 1 space kg comma space left parenthesis straight x subscript 3 comma straight y subscript 3 right parenthesis equals left parenthesis 0 comma 4 right parenthesis
straight X subscript CM equals fraction numerator straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 plus straight m subscript 3 straight x subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction equals space fraction numerator 1 cross times 0 plus 1 cross times 3 plus 1 cross times 0 over denominator 1 plus 1 plus 1 end fraction equals 1
straight Y subscript CM space equals space fraction numerator straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 plus straight m subscript 3 straight y subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction space equals space fraction numerator 1 cross times 0 plus 1 cross times 0 plus 1 cross times 4 over denominator 1 plus 1 plus 1 end fraction equals 4 over 3

The coordinates of centre of mass are open parentheses 1 comma 4 over 3 close parentheses .


Q.7. Particles of masses m, 2m, 3m... , nm gram are placed on the same line at distances l, 2l, 3l..., nl cm from a fixed point . The distance of centre of mass of the particles from the fixed point in centimeters is 

A) fraction numerator left parenthesis 2 straight n plus 1 right parenthesis straight l over denominator 3 end fraction 

B) fraction numerator straight l over denominator straight n plus 1 end fraction

C) fraction numerator straight n left parenthesis straight n squared plus 1 right parenthesis straight l over denominator 2 end fraction

D) fraction numerator 2 straight l over denominator straight n left parenthesis straight n squared plus 1 right parenthesis end fraction A) 140 m

B) 40 m 

C) 100 m 

D) 360 m 

Explanation-

straight X subscript CM equals fraction numerator straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 plus... plus straight m subscript straight n straight x subscript straight n over denominator straight m subscript 1 plus straight m subscript 2 plus... plus straight m subscript straight n end fraction
space space space space space space space space space equals fraction numerator ml plus 2 straight m times 2 straight l plus 3 straight m times 3 straight l plus... plus nm. nl over denominator straight m plus 2 straight m plus 3 straight m plus... nm end fraction
equals fraction numerator ml left parenthesis 1 plus 4 plus 9 plus... plus straight n squared right parenthesis over denominator straight m left parenthesis 1 plus 2 plus 3 plus... plus straight n right parenthesis end fraction space equals space fraction numerator straight l begin display style fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis 2 straight n plus 1 right parenthesis over denominator 6 end fraction end style over denominator begin display style fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction end style end fraction equals fraction numerator straight l left parenthesis 2 straight n plus 1 right parenthesis over denominator 3 end fraction 


Q.8. A train moving with a speed of 36 km/h takes 14 s to cross a bridge of length 100 m. The length of the train is 

A) 140 m

B) 40 m 

C) 100 m 

D) 360 m 

Explanation-

Speed of train , v=36 km/h =10 m/s Let x meter be the length of train. 

Total distance to be travelled by train to cross the bridge = (x+100) m 

As distance = speed cross timestime 

therefore (x+100)=10 cross times14 or x=40 m 


Q.9. The position of center of mass of a system of particles does not depend upon 

A) masses of particles

B) forces on particles 

C) position of the particles 

D) relative distance between the particles 

Explanation-

The position of center of mass of a system of particles does not depend upon the forces on the particles.


Q.10. Three particles of masses 1 kg, 2kg, and 3kf are situated at the corners of an equilateral triangle of side b in the x-y plane with mass 1kg  at the origin and 2 kg, on the x-axis. The coordinates of the center of mass are 

Aopen parentheses fraction numerator 7 straight b over denominator 12 end fraction comma fraction numerator 3 square root of 3 straight b end root over denominator 12 end fraction close parentheses

B) open parentheses fraction numerator 3 square root of 3 straight b end root over denominator 12 end fraction comma fraction numerator 7 straight b over denominator 12 end fraction close parentheses

C) open parentheses b over 12 comma fraction numerator 3 square root of 3 b end root over denominator 12 end fraction close parentheses

D) open parentheses fraction numerator 7 straight b over denominator 12 end fraction comma space fraction numerator square root of 3 straight b end root over denominator 12 end fraction close parentheses 

Explanation-

The coordinates of points A,B and C are (0,0) (b, 0) and open parentheses b over 2 comma space fraction numerator b square root of 3 over denominator 2 end fraction close parentheses respectively. 

straight X subscript CM equals fraction numerator 1 cross times 0 plus 2 cross times straight b plus 3 left parenthesis straight b divided by 2 right parenthesis over denominator 1 plus 2 plus 3 end fraction equals fraction numerator 7 straight b over denominator 12 end fraction
straight Y subscript CM space equals space fraction numerator 1 cross times 0 plus 2 cross times 0 plus 3 square root of 3 space left parenthesis straight b divided by 2 right parenthesis over denominator 1 plus 2 plus 3 end fraction equals space fraction numerator 3 square root of 3 straight b end root over denominator 12 end fraction

So, the coordinates of center of mass are 

open parentheses fraction numerator 7 straight b over denominator 12 end fraction comma fraction numerator 3 square root of 3 straight b end root over denominator 12 end fraction close parentheses