Dear Student, Soln (a): h=ut+12gt2For free fall: initial velocity, u = 0h = 12gt2g =2ht2=2×55×5=0 4m/s2 Soln b: Gravitational force on a body of mass m due to planet A: GMAmRA2 = mg Gravitational force on a body of mass m due to planet B: GMBmRB2 = mg' Given, RA = RB2 For g' = g2GMBRB2 = 12GMARA2 = 42GMARB2∴ MB = 2MA Soln (c): Mass on earth, m = 5 kg Weight on earth, w = mg = 49N ∴ g = 9 8 m/s2 gmoon=16×g=16×9 8 =1 63 m/s2 mass on moon will remain same i e 5 kg weight on moon = 16×49 = 8 17N

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Answer this question this 9 e 5. Solve the following exampleS• a. An ob_jcct takes 5 s to reach from a height of S m on a plan What is the value of g on the plonet2 Ans: 0.4 Inls2 b. The radius of planet A is half the radius of planet B. Ifthe mass of A IS MA, what must be the mass of B SO that the value of g on B is half that of its value on A? Ans: 2 MA c. The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 116th of that on the earth. Ans: 5 kg and 8.17 She tyre gravit the Use P roiect: Toke Find out moon $.06

Dear Student,
Soln (a):

h = u t + \frac{1}{2} g t^{2} F o r f r e e f a l l : i n i t i a l v e l o c i t y, u = 0 h = \frac{1}{2} g t^{2} g = \frac{2 h}{t^{2}} = \frac{2 \times 5}{5 \times 5} = 0.4 m / s^{2}

Soln b:
Gravitational force on a body of mass m due to planet A:

G \frac{M_{A} m}{{R_{A}}^{2}} = m g

Gravitational force on a body of mass m due to planet B:

G \frac{M_{B} m}{{R_{B}}^{2}} = m g'

Given,

R_{A} = \frac{R_{B}}{2}

For

g' = \frac{g}{2} \frac{G M_{B}}{{R_{B}}^{2}} = \frac{1}{2} \frac{G M_{A}}{{R_{A}}^{2}} = \frac{4}{2} \frac{G M_{A}}{{R_{B}}^{2}} ∴ M_{B} = 2 M_{A}

Soln (c):

Mass on earth, m = 5 kg
Weight on earth, w = mg = 49N

∴ g = 9.8 m / s^{2}

g_{m o o n} = \frac{1}{6} \times g = \frac{1}{6} \times 9.8 = 1.63 m / s^{2}

mass on moon will remain same i.e. 5 kg
weight on moon =

\frac{1}{6} \times 49 = 8.17 N

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