Solving Motion and Force Numericals

In the previous chapter, "The Three Equations of Motion," we armed ourselves with the mathematical tools needed to predict exactly how things move. We also know from earlier chapters that Force equals mass times acceleration (F = ma), and Weight equals mass times gravity (W = mg).

Knowing the formulas is one thing; knowing exactly how to apply them to a word problem under exam pressure is another. Many students panic when they see a paragraph full of numbers, vehicles, and moving objects. But physics numericals are actually the most reliable questions on your paper. Unlike theory questions where two options might look similar, a numerical has one exact, undeniable answer. If you follow the correct steps, you will find it every single time.

This chapter is entirely practical. We are going to take everything you have learned about motion, force, mass, and weight, and use it to solve standard exam problems step-by-step.

The Four-Step Master Method

Every single motion and force problem can be cracked using the same four-step sequence. Do not try to do the math in your head, and do not immediately start plugging numbers into a calculator. Follow this exact flow:

Let us break down exactly how to execute these steps before we dive into the actual problems.

Step 1: Decoding the English (Extracting Clues)

Examiners rarely give you the numbers directly. They hide them inside everyday English phrases. You have to translate the story into physics variables: Initial Velocity (u), Final Velocity (v), Acceleration (a), Distance/Displacement (s), and Time (t).

• "A car starts from rest..."

This is a hidden clue. It means the object was not moving at the beginning. Translation: Initial velocity (u) = 0.

• "...brakes are applied to bring the train to a halt."

This means the object eventually stopped moving. Translation: Final velocity (v) = 0.

• "A stone is dropped from a bridge..."

When something is dropped, it starts from a standstill, and gravity pulls it down. Translation: Initial velocity (u) = 0, and acceleration (a) = 9.8 m/s² (or 10 m/s² if the exam specifies).

• "...moves with a uniform velocity of 40 m/s."

If the velocity is uniform (constant), the speed is not changing. Translation: Acceleration (a) = 0.

Step 2: The Unit Conversion Trap

This is where 80% of students lose their marks. Physics formulas only work if all your ingredients are in the same language. That language is the SI System (International System of Units).

Before you calculate anything, ensure your data matches this checklist:

• Distance (s) must be in meters (m).
• Time (t) must be in seconds (s).
• Mass (m) must be in kilograms (kg).
• Velocity (u and v) must be in meters per second (m/s).

Step 3: Choosing the Right Formula

Write down the variables you have, and the one variable you need to find. Then, pick the formula that contains exactly those variables.

Here is your toolkit:

1. 1v = u + at (Use this when you don't know the distance, s)
2. 2s = ut + 1⁄2at² (Use this when you don't know the final velocity, v)
3. 3v² - u² = 2as (Use this when you don't know the time, t)
4. 4F = ma (Use this to connect motion to force)
5. 5W = mg (Use this to find weight from mass)

Worked Example 1: Calculating Final Velocity

Let us start with a straightforward kinematics problem to practice our four-step method.

The Problem: A motorcycle initially traveling at 36 km/h accelerates uniformly at a rate of 2 m/s² for 5 seconds. What is the final velocity of the motorcycle in m/s?

Step 1: Extract the clues

• Initial velocity (u) = 36 km/h
• Acceleration (a) = 2 m/s²
• Time (t) = 5 s
• Final velocity (v) = ? (This is what we need to find)

Step 2: Check the units Look at our initial velocity. It is in km/h. We must convert it to m/s. u = 36 × (5/18) = 10 m/s. Now our units are clean: u is in m/s, a is in m/s², and t is in seconds.

Step 3: Choose the formula We have u, a, and t. We need v. We do not care about distance (s). The perfect formula is the first equation of motion: v = u + at.

Step 4: Calculate

• v = 10 + (2 × 5)
• v = 10 + 10
• v = 20 m/s

Answer: The final velocity of the motorcycle is 20 m/s.

Worked Example 2: The Braking Vehicle (Retardation)

This is a classic exam favorite. It tests whether you remember that slowing down means negative acceleration.

The Problem: A train is traveling at a steady speed of 90 km/h. The driver applies the brakes, creating a uniform retardation of 0.5 m/s². How far will the train travel before coming to a complete halt?

Step 1: Extract the clues

• Initial velocity (u) = 90 km/h
• Retardation = 0.5 m/s².

(Wait! Retardation is negative acceleration. Therefore, a = -0.5 m/s²)

• "...coming to a complete halt." This means Final velocity (v) = 0.
• Distance (s) = ?

Step 2: Check the units The initial velocity is in km/h. Convert it! u = 90 × (5/18) = 25 m/s.

Step 3: Choose the formula We have u, v, and a. We need s. We do not know the time (t), and we don't need to find it. The equation that does not require time is the third equation of motion: v² - u² = 2as.

Step 4: Calculate

• 0² - (25)² = 2 × (-0.5) × s
• 0 - 625 = -1 × s
• -625 = -s
• s = 625 meters

Answer: The train will travel 625 meters before coming to a complete stop.

Worked Example 3: Finding the Stopping Force

Now we will combine the equations of motion with Newton's Second Law of Motion (F = ma). This is a two-step problem: first, we find the acceleration, and then we use it to find the force.

The Problem: A car of mass 1200 kg is moving at 72 km/h. The driver sees a roadblock and slams on the brakes, bringing the car to a complete stop in 4 seconds. Calculate the braking force exerted by the road on the tires.

Step 1: Extract the clues

• Mass (m) = 1200 kg
• Initial velocity (u) = 72 km/h
• Final velocity (v) = 0 (because it comes to a complete stop)
• Time (t) = 4 s
• Force (F) = ?

Step 2: Check the units Mass is in kg (good). Time is in seconds (good). Initial velocity is in km/h (bad). Convert it: u = 72 × (5/18) = 20 m/s.

Step 3: Choose the formula We need Force. The formula for force is F = ma. We have the mass (m = 1200), but we do not have the acceleration (a). We must find a first using our motion formulas. We have u, v, and t. We will use v = u + at to find a.

Step 4: Calculate Part A: Find the acceleration

• v = u + at
• 0 = 20 + a(4)
• Move 20 to the other side: -20 = 4a
• Divide by 4: a = -20 / 4
• a = -5 m/s²

(The negative sign makes sense because the car is slowing down).

Part B: Find the force

• F = ma
• F = 1200 × (-5)
• F = -6000 Newtons (N)

Answer: The braking force exerted on the car is -6000 N.

Worked Example 4: The Bullet and the Block (Advanced)

This problem tests your ability to handle very small numbers, very large numbers, and strict unit conversions all at once.

The Problem: A bullet of mass 20 grams traveling horizontally at 150 m/s strikes a stationary wooden block and comes to rest in 0.03 seconds. Calculate the magnitude of the force exerted by the wooden block on the bullet.

Step 1: Extract the clues

• Mass (m) = 20 grams
• Initial velocity (u) = 150 m/s
• Final velocity (v) = 0 (it comes to rest inside the block)
• Time (t) = 0.03 s
• Force (F) = ?

Step 2: Check the units Velocity is in m/s (good). Time is in seconds (good). Mass is in grams (trap!). We must convert it to kilograms. m = 20 / 1000 = 0.02 kg.

Step 3: Choose the formula Just like the previous example, we need F = ma. We have m, but we need to find a first using v = u + at.

Step 4: Calculate Part A: Find the acceleration

• v = u + at
• 0 = 150 + a(0.03)
• -150 = 0.03a
• a = -150 / 0.03
• a = -5000 m/s²

(Bullets slow down incredibly fast, so a massive negative acceleration is correct).

Part B: Find the force

• F = ma
• F = 0.02 × (-5000)
• F = -100 N

Answer: The magnitude of the force exerted by the block on the bullet is 100 N.

Worked Example 5: Weight on Different Planets

Let us shift away from moving vehicles and look at gravity. Remember the golden rule from the previous chapters: Mass never changes, but weight changes depending on gravity.

The formula for weight is a specific version of Newton's Second Law (F=ma). Instead of general force (F), we use Weight (W). Instead of general acceleration (a), we use the acceleration due to gravity (g). Therefore, W = mg.

The Problem: An astronaut has a mass of 80 kg on Earth. The acceleration due to gravity on Earth is 9.8 m/s². The astronaut travels to Mars, where the acceleration due to gravity is 3.7 m/s². Calculate the astronaut's weight on Earth, their mass on Mars, and their weight on Mars.

Step 1 & 2: Extract clues and check units

• Mass (m) = 80 kg (Unit is correct)
• Gravity on Earth (g_earth) = 9.8 m/s²
• Gravity on Mars (g_mars) = 3.7 m/s²

Step 3 & 4: Calculate

Question 1: Weight on Earth

• W = m × g_earth
• W = 80 × 9.8
• W = 784 N

Question 2: Mass on Mars

• This is a trick question! Mass is the amount of matter in an object. It does not change no matter where you are in the universe.
• Mass on Mars = 80 kg.

Question 3: Weight on Mars

• W = m × g_mars
• W = 80 × 3.7
• W = 296 N

Quick Revision

Scan these high-yield points before your exam to ensure you don't fall for common numerical traps:

• The 5/18 Rule: To convert km/h to m/s, multiply by 5/18. (e.g., 72 km/h = 20 m/s).
• The Gram Trap: Always convert mass from grams to kilograms by dividing by 1000 before using F=ma or W=mg.
• Hidden Zeros: "Starts from rest" means u = 0. "Comes to a stop" means v = 0.
• Constant Speed: If an object moves at a uniform/constant velocity, its acceleration is exactly 0.
• Retardation is Negative: If brakes are applied or an object is slowing down, the acceleration value must be written with a minus sign (e.g., -2 m/s²).
• Negative Force: A negative result for Force simply means the force is acting in the opposite direction to the object's movement (like friction or brakes).
• Mass vs. Weight: Mass (kg) is constant everywhere in the universe. Weight (N) changes depending on local gravity (W=mg).
• Choosing Equations: If you don't have time (t) and don't need it, use v² - u² = 2as. If you don't have distance (s), use v = u + at.