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Byju's Answer

Standard IX

Physics

Introduction to Force

Two masses ...

Question

Two masses $m_{1}$ & $m_{2}$ are placed $r$ distance apart as shown in figure. $(m_{1} \neq m_{2})$
$F_{1}$ and $F_{2}$ are gravitational forces acting on $m_{1}$ and $m_{2}$ respectively.
$a_{1}$ and $a_{2}$ are acceleration of $m_{1}$ and $m_{2}$ respectively.
Choose correct option-
a) $(_{1} =_{2})$ e) $(_{1} = -_{2})$
b) $(_{1} = -_{2})$ f) $(a_{C.M} = 0)$
c) $(F_{net} = 0)$ g) $(v_{C.M} = Constant)$
d) $(_{1} =_{2})$
[ $a_{C.M}$ is acceleration of Center of Mass]
[ $v_{C.M}$ is velocity of Center of Mass]

a, c, f

are correct

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b, c, e, f

are correct

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c, d, f, g

are correct

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b, c, f, g

are correct

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Solution

The correct option is C $b, c, f, g$ are correct
$_{1} = \frac{G m_{1} m_{2}}{r^{2}} (^i)$

$_{2} = \frac{G m_{1} m_{2}}{r^{2}} (-^i)$

So, $_{1} \neq_{2}$
$_{1} = -_{2}$ or $F_{net} = 0$

If $F_{net} = 0$ , then $a_{C . M} = 0$
If $a_{C . M} = 0$ , then $v_{C . M} =$ Constant

Now, $a_{1} = \frac{F_{1}}{m_{1}} = \frac{G m_{2}}{r^{2}}$ , $a_{2} = \frac{F_{2}}{m_{2}} = \frac{G m_{1}}{r^{2}}$
So, $a_{1} \neq a_{2}$

b, c , f , g are correct

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The correct option is C b,c,f,g are correct→F1=Gm1m2r2(^i)→F2=Gm1m2r2(−^i) So, →F1≠→F2 →F1=−→F2 or Fnet=0 If Fnet=0, then aC.M=0If aC.M=0, then vC.M= Constant Now, a1=F1m1=Gm2r2, a2=F2m2=Gm1r2 So, a1≠a2 b, c , f , g are correct