Solved Problems on Laws of Motion - Science

SOLVED PROBLEMS

Problem-1

Calculate the velocity of a moving body of mass 5 kg whose linear momentum is 2.5 kg m s^–1.

Solution:

Linear momentum = mass × velocity

Velocity = linear momentum / mass. V = 2.5 / 5 = 0.5 m s^–1

Problem 2

A door is pushed, at a point whose distance from the hinges is 90 cm, with a force of 40 N. Calculate the moment of the force about the hinges.

Solution:

Formula: The moment of a force M = F × d

Given: F = 40 N and d = 90 cm = 0.9 m.

Hence, moment of the force = 40 × 0.9 = 36 N m.

Problem 3

At what height from the centre of the Earth the acceleration due to gravity will be 1/4^th of its value as at the Earth.

Solution:

Data: Height from the centre of the Earth, Rʹ = R + h

The acceleration due to gravity at that height, gʹ = g/4

From the centre of the Earth, the object is placed at twice the radius of the earth.

The ratio of masses of two planets is 2:3 and the ratio of their radii is 4:7 Find the ratio of their accelerations due to gravity. - Science

Cosmic Calculations: How Strong is Gravity on Other Worlds?

We all feel gravity's pull every day, but have you ever wondered how it might differ on other planets? The "weight" we feel is determined by a planet's acceleration due to gravity, a value that depends on both its mass and its size.

A more massive planet will have a stronger gravitational pull, but a larger radius will weaken that pull at the surface. Understanding this balance allows us to compare different worlds, even without visiting them. Let's explore this with a fascinating problem.

The Problem

The ratio of the masses of two planets is 2:3 and the ratio of their radii is 4:7.

Question: Find the ratio of their accelerations due to gravity.

The Science Behind the Solution

The acceleration due to gravity, denoted by '$g$', is calculated using Newton's law of universal gravitation. The formula is:

$$g = \frac{GM}{R^2}$$

Here, '$G$' is the gravitational constant, '$M$' is the mass of the planet, and '$R$' is its radius. Since we are looking for the ratio of the gravities of two planets (let's call them Planet 1 and Planet 2), we can set up the following relation:

$$\frac{g_1}{g_2} = \frac{GM_1/R_1^2}{GM_2/R_2^2}$$

The gravitational constant '$G$' cancels out, simplifying the equation to:

$$\frac{g_1}{g_2} = \frac{M_1}{M_2} \times \left(\frac{R_2}{R_1}\right)^2$$

Solving the Problem Step-by-Step

Let's list what we know from the problem:

• Ratio of masses ($M_1 : M_2$): 2 : 3, which means $\frac{M_1}{M_2} = \frac{2}{3}$
• Ratio of radii ($R_1 : R_2$): 4 : 7, which means $\frac{R_1}{R_2} = \frac{4}{7}$

Notice that our formula needs the ratio $\frac{R_2}{R_1}$. We can easily find this by inverting the ratio we have: if $\frac{R_1}{R_2} = \frac{4}{7}$, then $\frac{R_2}{R_1} = \frac{7}{4}$.

Now, we plug these ratios into our simplified formula:

$$\frac{g_1}{g_2} = \frac{2}{3} \times \left(\frac{7}{4}\right)^2$$

First, we square the radii ratio:

$$\frac{g_1}{g_2} = \frac{2}{3} \times \frac{49}{16}$$

Finally, we multiply the fractions:

$$\frac{g_1}{g_2} = \frac{2 \times 49}{3 \times 16} = \frac{98}{48}$$

This fraction can be simplified by dividing the numerator and denominator by 2:

$$\frac{g_1}{g_2} = \frac{49}{24}$$

Answer: The ratio of the accelerations due to gravity ($g_1 : g_2$) is 49:24.

The Takeaway

This result shows how both mass and radius play a crucial role. Even though Planet 2 is more massive (3/2 times the mass of Planet 1), its much larger radius means its surface gravity is actually weaker. It's a perfect example of how cosmic properties are about more than just one factor!