Question

Two particals of mass 200 g each are placed at a separastion of 10 cm. Assume that the only force acting on them are due to their gravitational attraction.Find the acceleration of each when they are allowed to move.

Answer 1

Given

• Mass of two particles = 200 g
• Distance = 10 cm

To Find

• Gravitational Acceleration when they are allowed to move

Solution

• Here we o ought to first convert the units. Mass of the bodies are to be converted from g → Kg and the distance from cm → m

Converting Units

→ Mass = 200 g

→ Mass = 200/1000

→ Mass = 0.2 kg

Similarly,

→ Distance = 10 cm

→ Distance = 10/100

→ Distance = 0.1 m

━━━━━━━━━━━━━━

✭ Gravitational Force :

→ F = (Gm₁m₂)/r²

→ F = (6.67×10⁻¹¹ × 0.2 × 0.2)/0.1²

→ F = (6.67×10⁻¹¹ × 0.2²)/0.01

→ F = (6.67×10⁻¹¹ × 0.04)/0.01

→ F = 2.68 × 10⁻¹¹

━━━━━━━━━━━━━━

• Now using Newton's second law of motion we may find the acceleration of the body

✭ Acceleration of the body :

→ F = ma

→ 2.68 × 10⁻¹¹ = 0.2 × a

→ 2.68 × 10⁻¹¹/0.2 = a

→ Acceleration = 1.34 × 10⁻⁹

Answer 2

$\huge \rm \: given$

• Mass of particles = 200 g
• Separation between them = 10 cm

$\huge \rm \: to \: find$

acceleration of each when they are allowed to move.

$\huge \rm \: solution$

In this case we will convert

${ \sf {\blue { g \longrightarrow \: kg}}}$

$\sf \green {cm \longrightarrow \: m}$

$\rm \: 200 \: g \: = \dfrac{200}{1000} \: kg$

$\rm \: 200 \: g \: = 0.2 \: kg$

$\rm \: 10 \: cm \: = \dfrac{10}{100} cm$

$\rm \: 10 \: cm \: = 0.1 \: m$

First we will find gravitational force

$\huge \bf \red \bigstar \: Gravitational \: force$

$\sf \: f \: = \dfrac{(gm1m2)} {r}{2}$

F = (6.67×10⁻¹¹ × 0.2 × 0.2)/0.1²

F = (6.67×10⁻¹¹ × 0.2²)/0.01

F = (6.67×10⁻¹¹ × 0.04)/0.01

F = 2.68 × 10⁻¹¹

Now here we will apply Newton's second law of motion.

$\huge \fbox {f \: = ma}$

$\sf 2.68 × 10⁻¹¹ = 0.2 \times a$

$\sf \: \: a \: = \frac{2.68 \times {10}^{ - 11} }{0.2}$

$\huge \bf \: acceleration = 1.34 \times {10}^{ - 9}$