Problem 15 - Master Resource Book in JEE Main Physics

Question

A body projected with velocity \(u\) at projection angle \(\theta\) has horizontal range \(R\). For the same velocity and projection angle, its range on the moon surface will be \(g_{\text {moen }}=g_{\text {earth }} / 6\) ) (a) \(36 R\) (b) \(\frac{R}{36}\) (c) \(\frac{R}{16}\) (d) \(6 R\)

Step-by-Step Solution

Answer

(d) The range on the moon is 6R.

Step 1: Understanding Base Scenario

For a projectile on Earth, the horizontal range is given by the formula \( R = \frac{u^2 \sin(2\theta)}{g_{\text{earth}}} \), where \( u \) is the initial velocity and \( \theta \) is the projection angle, and \( g_{\text{earth}} \) is the acceleration due to gravity on Earth.

Step 2: Physics on the Moon

On the moon, the acceleration due to gravity \( g_{\text{moon}} \) is \( \frac{g_{\text{earth}}}{6} \). Therefore, the formula for the horizontal range \( R_{\text{moon}} \) becomes \( R_{\text{moon}} = \frac{u^2 \sin(2\theta)}{g_{\text{moon}}} \).

Step 3: Substituting Moon's Gravity

Substitute \( g_{\text{moon}} = \frac{g_{\text{earth}}}{6} \) into the range formula: \[ R_{\text{moon}} = \frac{u^2 \sin(2\theta)}{\frac{g_{\text{earth}}}{6}} = \frac{6u^2 \sin(2\theta)}{g_{\text{earth}}} \].

Step 4: Relating to Earth Range

Notice that \( \frac{u^2 \sin(2\theta)}{g_{\text{earth}}} = R \), which is the range on Earth. Therefore, \[ R_{\text{moon}} = 6 \times R \].

Step 5: Final Answer

Therefore, the range on the moon is 6 times the range on Earth. The correct choice is (d) \( 6R \).

Key Concepts

Range of a Projectile

Projectile motion involves analyzing how an object moves through two dimensions - horizontally and vertically. A crucial factor in this is determining the range, which is the horizontal displacement covered by the projectile.
The formula for the range of a projectile fired with an initial velocity \( u \) at an angle \( \theta \) on Earth is given by:\[R = \frac{u^2 \sin(2\theta)}{g_{\text{earth}}}\]Here:

• \( u \) is the initial velocity
• \( \theta \) is the angle of projection
• \( g_{\text{earth}} \) is the acceleration due to gravity on Earth

This formula shows that the range depends on both the initial speed and the angle of launch. Maximizing the range involves launching at an appropriate angle, typically 45 degrees to maximize the \( \sin(2\theta) \) term.
When considering other celestial bodies like the moon, you need to account for how different gravity affects projectile range.

Gravity on Moon

The gravitational pull on the moon is a lot weaker than on Earth. Specifically, it's one-sixth of Earth's gravity.
This significant difference can dramatically alter the behavior of projectiles. On the moon, where the acceleration due to gravity \( g_{\text{moon}} \) is \( \frac{g_{\text{earth}}}{6} \), the range formula for a projectile becomes:\[R_{\text{moon}} = \frac{u^2 \sin(2\theta)}{g_{\text{moon}}}\]When diving into the math:

• Substituting \( g_{\text{moon}} = \frac{g_{\text{earth}}}{6} \) into the equation
• Leads to \( R_{\text{moon}} = \frac{6u^2 \sin(2\theta)}{g_{\text{earth}}} \)
• This shows that the range on the moon is six times that on Earth
  

This highlights the profound impact that gravity has on projectile motion.

Physics Problem Solving

Solving physics problems, especially those involving projectile motion, requires a structured approach. Here's how you can tackle these problems:
**Identify Given Variables and Formulas**

• Start by listing known values like initial velocity \( u \), angle \( \theta \), and local gravity \( g \).
• Recognize which gravitational environment (Earth vs Moon) you are dealing with.

**Choose Suitable Equations**

• Use the proper equation: \( R = \frac{u^2 \sin(2\theta)}{g} \) based on whether you're on Earth or another celestial body.

**Substitute and Solve for Range**

• Plug known values into the equation, simplifying where possible.
• Check your calculations carefully to ensure accuracy.

Following this method comprehensively guides your problem-solving and ensures you don't miss critical steps, leading to a more confident understanding of the physics involved.