Newton's Third Law Problems- Practice and Solutions
What Newton's Third Law Actually Means
For every action, there is an equal and opposite reaction. You've heard it. You've probably memorized it. But most students still get these problems wrong.
The law is simple on paper. When two objects interact, the force object A exerts on object B equals the force object B exerts on object A. Same magnitude. Opposite direction. That's it.
The confusion comes from thinking these forces cancel out. They don't. Action-reaction pairs act on different objects. That's the part most people miss.
The Key Distinction You Need to Make
Consider a book sitting on a table. Gravity pulls down on the book. The table pushes up on the book. Are these Newton's third law pairs?
No. They're not.
Here's why: Newton's third law says the forces must be between two different objects, and they must be of the same type. Gravity is a field force. The normal force from the table is a contact force. Different types.
The actual third law pair is: the book pulls Earth upward with the same force Earth pulls the book downward. The table pushes the book upward, and the book pushes the table downward. Those are action-reaction pairs.
How to Approach Newton's Third Law Problems
Most problems give you a scenario and ask you to identify forces or calculate accelerations. Here's your step-by-step process:
- Identify all objects in the system
- Draw free body diagrams for each object separately
- For each force, ask: "What object is causing this force?"
- Find the reaction force on the other object
- Apply Newton's Second Law (F = ma) to each object
- Solve the system of equations
The mistake students make is drawing one diagram for the whole system. Don't do that. Every object gets its own diagram.
Practice Problems with Solutions
Problem 1: The Basic Push
A 5 kg block sits on a frictionless table. You push on the block with a horizontal force of 20 N. What is the acceleration of the block? What force does the block exert on you?
Solution:
The block accelerates because of the net force acting on it. Using F = ma:
a = F/m = 20 N / 5 kg = 4 m/s²
Newton's third law tells us the block pushes back on you with the same magnitude: 20 N in the opposite direction. Your acceleration depends on the friction between your feet and the floor—if you're well-anchored, you won't move much. But the force is real and equal.
Problem 2: Two Blocks Together
A 3 kg block rests on top of a 7 kg block. The 7 kg block sits on a frictionless table. You pull the 7 kg block to the right with force F = 50 N. What is the acceleration of each block? What is the force between them?
Solution:
First, analyze the top block (3 kg). The only horizontal force on it is the friction force from the bottom block pushing it forward. Call this f. Using F = ma:
f = 3a
Now the bottom block. It has the applied force 50 N pushing it right, and the top block pushes it left with force f (Newton's third law—the top block pushes back on the bottom block).
50 - f = 7a
Substitute f = 3a:
50 - 3a = 7a
50 = 10a
a = 5 m/s²
The force between blocks: f = 3 × 5 = 15 N
The bottom block accelerates at 5 m/s². The top block also accelerates at 5 m/s² (they move together). The force transmitted between them is 15 N.
Problem 3: The Tension Problem
A 4 kg hanging mass connects to a 6 kg mass on a frictionless table via a light rope over a pulley. What is the acceleration of the system? What is the tension in the rope?
Solution:
Draw two free body diagrams: one for the hanging mass, one for the table mass.
For the hanging mass (taking downward as positive):
mg - T = ma
40 - T = 4a
For the table mass (taking right as positive):
T = 6a
Substitute T = 6a into the first equation:
40 - 6a = 4a
40 = 10a
a = 4 m/s²
T = 6a = 24 N
Notice the tension is not half the weight. It's less because the hanging mass has to accelerate. If the masses were equal and both hanging, tension would equal mg/2.
Problem 4: Inclined Plane with a Twist
A 2 kg block slides down a frictionless 30° incline. At the base, it hits and sticks to a stationary 4 kg block. The two blocks stick together. How far do they slide before stopping if the coefficient of friction is 0.3?
Solution:
First, find the velocity of the 2 kg block when it reaches the bottom.
Using conservation of energy (or kinematics with acceleration g·sinθ):
v² = 2(g·sin30°)d
Where d is the length of the incline. We need the height h = d·sin30° instead. The block falls height h, so:
v² = 2gh = 2(10)(h) = 20h
After collision, momentum conserved (inelastic collision):
(2 kg)(v) = (6 kg)(v')
v' = v/3
Now the combined 6 kg mass slides and friction acts. The friction force: f = μN = 0.3 × (6 × 10 × cos30°) = 0.3 × 52 = 15.6 N
Using work-energy: work done by friction = change in kinetic energy
(15.6)(d) = ½(6)(v'²)
Substitute v'² = v²/9 = (20h)/9
15.6d = 3 × (20h/9) = 20h/3
If the incline was height h, the distance along the plane was h/sin30° = 2h. So d = 2h.
15.6(2h) = 20h/3
31.2h = 20h/3
31.2h = 6.67h
This gives a negative result—the blocks would actually continue moving in the direction of the incline. The problem setup matters here. For a flat surface after the collision, solve using the actual distance given or the velocity found.
The point: break problems into stages. Collision analysis uses momentum. Sliding analysis uses energy or forces. Don't mix them up.
Force Comparison Table
| Scenario | Forces on Object A | Reaction Force on Object B | Net Force on A? |
|---|---|---|---|
| Book on table | Table pushes up on book | Book pushes down on table | Yes (if forces unbalanced) |
| Person pushing wall | Wall pushes back on person | Person pushes on wall | Yes (person accelerates backward) |
| Two blocks in contact | Block 2 pushes on block 1 | Block 1 pushes on block 2 | Depends on other forces |
| Rocket in space | Exhaust gases push rocket forward | Rocket pushes exhaust backward | Yes (rocket accelerates) |
| Earth-book gravity | Earth pulls book down | Book pulls Earth up | Yes (book falls) |
Common Mistakes That Cost You Points
- Identifying forces as a third law pair when they're on the same object. The normal force and gravity on a book are both on the book. They're not an action-reaction pair.
- Forgetting that action-reaction forces have the same nature. Both must be contact forces, or both field forces. Gravity-gravity, normal-normal, tension-tension.
- Thinking equal and opposite means they cancel. They act on different objects. They never cancel each other.
- Skipping the free body diagram. If you don't draw separate diagrams for each object, you'll mix up your forces every time.
Quick Reference: Newton's Third Law Checklist
When you see a force, ask these questions:
- What object is exerting this force?
- What object is receiving this force?
- Is there a corresponding force of the same type going the opposite direction?
- Are these two forces acting on the same object or different objects?
If the forces are on different objects and of the same type (both contact, both gravitational, etc.), you've found your action-reaction pair.
Practice with five problems, and you'll stop second-guessing yourself. These problems follow patterns. Once you see the pattern, you'll solve them in under a minute.