Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2 ) from Physics Gravitation Class 11 Manipur Board
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Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2 )


C.

The concept of relative motion can be applied to predict the nature of motion of one particle with respect to the other.

Consider the stones thrown up simultaneously as shown in the diagram below.

Considering the motion of the second particle with respect to the first we have relative acceleration
|a21| = |a2-a1| = g-g = 0



Thus, motion of the first particle is straight line with respect to the second particle till the first particle strikes ground at a time given by
negative 240 space equals space 10 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
or space straight t squared minus 2 straight t minus 48 space equals space 0
or space straight t squared space minus 8 straight t space plus space 6 straight t space minus 48 space equals space 0
or space space straight t squared space equals space 8 comma negative 6 space left parenthesis not space possible right parenthesis
Thus, distance covered by the second particle with respect to the first particle in 8s is
S12 = (v21) t = (40-10)(8s)
 = 30 x 8 = 240m
Similarly, time taken by the second particle to strike the ground is given by
negative 240 space equals space 40 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
minus 240 space equals space 40 space straight t space minus 5 straight t squared
5 straight t squared minus 40 straight t minus 240 space equals 0
straight t squared minus 8 straight t minus 48 space equals space 0
straight t squared minus 12 straight t space plus space 4 straight t minus 48 space equals 0
straight t left parenthesis straight t minus 12 right parenthesis plus 4 left parenthesis straight t minus 12 right parenthesis space equals 0
straight t space equals space 12 comma negative 4 space left parenthesis not space possible right parenthesis
Thus, after the 8s magnitude of relative velocity will increases up to 12 s when the second particle strikes the ground.

The concept of relative motion can be applied to predict the nature of motion of one particle with respect to the other.

Consider the stones thrown up simultaneously as shown in the diagram below.

Considering the motion of the second particle with respect to the first we have relative acceleration
|a21| = |a2-a1| = g-g = 0



Thus, motion of the first particle is straight line with respect to the second particle till the first particle strikes ground at a time given by
negative 240 space equals space 10 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
or space straight t squared minus 2 straight t minus 48 space equals space 0
or space straight t squared space minus 8 straight t space plus space 6 straight t space minus 48 space equals space 0
or space space straight t squared space equals space 8 comma negative 6 space left parenthesis not space possible right parenthesis
Thus, distance covered by the second particle with respect to the first particle in 8s is
S12 = (v21) t = (40-10)(8s)
 = 30 x 8 = 240m
Similarly, time taken by the second particle to strike the ground is given by
negative 240 space equals space 40 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
minus 240 space equals space 40 space straight t space minus 5 straight t squared
5 straight t squared minus 40 straight t minus 240 space equals 0
straight t squared minus 8 straight t minus 48 space equals space 0
straight t squared minus 12 straight t space plus space 4 straight t minus 48 space equals 0
straight t left parenthesis straight t minus 12 right parenthesis plus 4 left parenthesis straight t minus 12 right parenthesis space equals 0
straight t space equals space 12 comma negative 4 space left parenthesis not space possible right parenthesis
Thus, after the 8s magnitude of relative velocity will increases up to 12 s when the second particle strikes the ground.

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The position co-ordinates of two particles of masses m1 and m2are (x1, y1, z1) and (x2, y2, z2) respectively. Find the coordinates of the centre of mass.

The position vectors of masses m1 and m2 are respectively,

space space space space space space space space space space space space r with rightwards harpoon with barb upwards on top subscript 1 equals space straight x subscript 1 straight i with overbrace on top plus straight y subscript 1 straight j with overbrace on top plus straight z subscript 1 straight k with overbrace on top

a n d space space space space space r with rightwards harpoon with barb upwards on top subscript 2 space equals space straight x subscript 2 straight i with overbrace on top plus straight y subscript 2 straight j with overbrace on top plus straight z subscript 2 straight k with overbrace on top

Let the position coordinates of the centre of mass be (X, Y, Z).

Therefore the position vector of centre of mass is,


R with rightwards harpoon with barb upwards on top equals straight X straight i with overbrace on top plus straight Y straight j with overbrace on top plus straight Z straight k with overbrace on top

Since,    R with rightwards harpoon with barb upwards on top equals fraction numerator straight m subscript 1 stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus straight m subscript 2 space straight r with rightwards harpoon with barb upwards on top subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space

therefore space space space straight X straight i with overbrace on top plus straight Y straight j with overbrace on top plus straight Z straight k with overbrace on top 

equals space space fraction numerator straight m subscript 1 left parenthesis straight x subscript 1 straight i with overbrace on top plus straight y subscript 1 straight j with overbrace on top plus straight z subscript 1 straight k with overbrace on top right parenthesis plus straight m subscript 2 left parenthesis straight x subscript 2 straight i with overbrace on top plus straight y subscript 2 straight j with overbrace on top plus straight z subscript 2 straight k with overbrace on top right parenthesis over denominator straight m subscript 1 plus straight m subscript 2 end fraction

space equals space space fraction numerator open parentheses straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 close parentheses over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight i with overbrace on top plus fraction numerator left parenthesis straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight j with overbrace on topplus fraction numerator left parenthesis straight m subscript 1 straight z subscript 1 plus straight m subscript 2 straight z subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space straight k with overbrace on top

Comparing the coefficients of  straight i with overbrace on top comma space straight j with overbrace on top space and space straight k with overbrace on top, we get

straight X equals fraction numerator left parenthesis straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
straight Y space equals space straight X equals fraction numerator left parenthesis straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
straight z equals fraction numerator left parenthesis straight m subscript 1 straight z subscript 1 plus straight m subscript 2 straight z subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction 


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