Momentum and Impulse

16.1 Why Momentum Matters

Momentum is a conserved quantity: the total momentum of an isolated system never changes, no matter how complex the internal interactions. This makes it a powerful tool for analyzing collisions, explosions, and interactions where forces are too brief or too variable to treat with Newton’s laws directly.

The work–energy theorem connects force and displacement to changes in kinetic energy. Momentum connects force and time to changes in motion. Together, these two perspectives — energy and momentum — give a complete picture of how interactions transfer both energy and motion between objects.

Momentum is particularly valuable when the collision time is very short and forces are not known precisely; when we know initial and final states but not the details of the interaction; and when multiple objects interact and we want to track the combined behavior of the system without analyzing every internal force.

16.2 Defining Linear Momentum

The linear momentum of an object is the product of its mass and its velocity:

p = mv

The SI unit of momentum is kg·m/s, which is equivalent to N·s. Both are correct and used in different contexts.

Newton’s second law can be restated in terms of momentum. The net force on an object equals the rate of change of its momentum:

F_net = Δp / Δt

16.3 Momentum as a Vector Quantity

Momentum is a vector: it has both magnitude and direction. The direction of momentum is the same as the direction of velocity. In one-dimensional problems, positive and negative signs encode direction. In two-dimensional problems, momentum must be analyzed by components.

The total momentum of a system of objects is the vector sum of the momenta of all individual objects:

p_total = p₁ + p₂ + p₃ + … = m₁v₁ + m₂v₂ + …

When two objects move in opposite directions, their momenta partially or fully cancel. When they move in the same direction, their momenta add. This vector nature is crucial for understanding collisions and conservation of momentum.

16.4 Impulse and Change in Momentum

When a net force acts on an object for a time interval Δt, the object’s momentum changes. The quantity that measures this change is called impulse.

Impulse is defined as the product of the average net force and the time interval over which it acts:

J = F_avg Δt

Impulse is particularly useful when the force is large and brief, as in a collision, making the average force a more practical quantity than the instantaneous force history.

Graphically, impulse is the area under a force-time graph. For a constant force, this is simply F × Δt. For a variable force, the area must be estimated.

16.5 The Impulse–Momentum Theorem

The impulse–momentum theorem states that the impulse acting on an object equals its change in momentum:

J = Δp = p_f − p_i = mv_f − mv_i

This theorem follows directly from Newton’s second law. It is useful because it connects observable quantities, such as initial and final velocities or force and time, without requiring knowledge of acceleration at every instant.

16.6 Force-Time Relationships

In a real collision, the force is not constant but varies with time: it rises from zero to a peak value and then falls back to zero as the objects deform and separate. The impulse is the area under this force-time curve.

For a given impulse, there is a trade-off between force and time: a larger contact time results in a smaller average force, and vice versa.

16.7 Why Increasing Collision Time Reduces Force

The impulse–momentum theorem J = F_avg Δt = Δp shows that for a fixed change in momentum, force and time are inversely related.

• Airbags extend the time over which a driver’s momentum changes during a crash.
• Gymnasts and martial artists bend their knees on landing to increase contact time.
• Crumple zones in cars deform progressively, increasing crash duration.
• Catching a ball with soft hands extends the stopping time and reduces force.
• Foam padding, helmets, and shock-absorbing soles all work by increasing Δt for a given Δp.

16.8 Momentum in Isolated Systems

A system is isolated, or closed, if no net external force acts on it. In an isolated system, the total momentum is constant, regardless of what happens inside the system.

Internal forces are forces between objects within the system; they always come in Newton’s third law pairs and cancel when summed over the whole system. External forces are forces from outside the system; they change the total momentum.

ΣF_external = 0 ⇒ p_total = constant

16.9 Conservation of Momentum

In an isolated system, the total momentum before an interaction equals the total momentum after the interaction:

p_total,i = p_total,f

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

Conservation of momentum is one of the most fundamental laws in physics. It holds in classical mechanics, quantum mechanics, special relativity, and nuclear physics.

16.10 Recoil and Propulsion

When two objects that are initially at rest push apart, momentum is conserved. The total momentum before was zero, so the total momentum after must also be zero. The momenta of the two objects are equal in magnitude and opposite in direction.

m₁v₁ + m₂v₂ = 0 ⇒ m₁v₁ = −m₂v₂

This is the principle behind recoil, rocket propulsion, and jet propulsion. The lighter object acquires greater speed, but the momenta are equal and opposite.

16.11 Explosions and Separation Events

An explosion is the reverse of a perfectly inelastic collision: a single object separates into multiple parts, with internal energy converted to kinetic energy. Momentum is still conserved. If the system starts at rest, the total final momentum is zero.

In an explosion, kinetic energy increases because internal energy is converted to kinetic energy. Total momentum is conserved, but kinetic energy is not.

16.12 Common Mistakes in Momentum Analysis

Chapter Summary

Momentum is the product of mass and velocity: p = mv. It is a conserved vector quantity in isolated systems. The impulse–momentum theorem connects force, time, and momentum change.

• Linear momentum: p = mv; SI unit kg·m/s = N·s; vector quantity.
• Newton’s second law in momentum form: F_net = Δp/Δt.
• Impulse: J = F_avgΔt = Δp; equals change in momentum and the area under an F-t graph.
• For fixed Δp, longer collision time means smaller average force.
• In an isolated system, total momentum before equals total momentum after.
• Elastic collisions conserve both momentum and kinetic energy. Inelastic collisions conserve only momentum. Explosions conserve momentum but increase kinetic energy.

Key Terms

linear momentum

The product of an object’s mass and velocity: p = mv; a vector quantity.

impulse

The product of average net force and the time interval over which it acts: J = F_avgΔt; equals change in momentum.

impulse–momentum theorem

The impulse acting on an object equals its change in momentum.

isolated system

A system on which the net external force is zero.

conservation of momentum

The total momentum of an isolated system remains constant.

elastic collision

A collision in which both momentum and kinetic energy are conserved.

inelastic collision

A collision in which momentum is conserved but kinetic energy decreases.

perfectly inelastic collision

An inelastic collision in which objects stick together and move with a common final velocity.

explosion

A separation event in which one object or system separates into parts; momentum is conserved but kinetic energy increases.

recoil

The backward motion of an object when it expels mass or exerts a force forward.

Concept Review Questions

1. Define linear momentum. What does it mean to say momentum is a vector quantity?
2. How is Newton’s second law related to the rate of change of momentum? Write the momentum form of Newton’s second law.
3. What is impulse? How is it related to the change in momentum of an object?
4. An egg is dropped from the same height onto a hard floor and onto a foam pad. The momentum change is the same in both cases. Why is the egg less likely to break on the foam pad?
5. What conditions must be satisfied for momentum to be conserved in a collision?
6. Two objects collide in an isolated system. One gains momentum. What happens to the other’s momentum? Explain using Newton’s third law.
7. Distinguish between elastic and inelastic collisions. Is kinetic energy conserved in either or both? Is momentum conserved in either or both?
8. A rifle fires a bullet. Explain the recoil of the rifle using conservation of momentum.
9. Two identical balls roll toward each other at the same speed. What is the total momentum of the system?
10. Can an object have momentum without kinetic energy? Can it have kinetic energy without momentum?

Momentum Problems

11. Find the momentum of: (a) a 0.145 kg baseball at 45.0 m/s; (b) a 70.0 kg sprinter at 10.0 m/s; (c) a 1200 kg car at 30.0 m/s; (d) a 0.010 kg bullet at 850 m/s.
12. A 60.0 kg person stands at rest on a frictionless ice rink and catches a 0.500 kg ball thrown at 12.0 m/s. Find the velocity of person+ball immediately after catching.
13. A 3.0 kg object moving at +6.0 m/s collides with a 5.0 kg object at rest. After the collision, the 3.0 kg object moves at +1.0 m/s. Find the velocity of the 5.0 kg object after.
14. A 1500 kg car moving at +20.0 m/s rear-ends a 2000 kg truck moving at +12.0 m/s in the same direction. They stick together. Find the combined velocity.
15. Two ice skaters, 60.0 kg and 80.0 kg, push off each other from rest. The 60.0 kg skater moves at 3.0 m/s to the right. Find the velocity of the 80.0 kg skater.

Impulse Problems

16. A force of 800 N acts on a 0.50 kg ball for 0.040 s. (a) Find the impulse. (b) If the ball was initially at rest, find its final speed.
17. A 0.145 kg baseball, pitched at 40.0 m/s, is hit and moves at +55.0 m/s. (a) Find the impulse from the bat. (b) If contact lasts 1.5 ms, find the average force.
18. A 70.0 kg runner is brought to rest from 8.0 m/s in 0.60 s. Find: (a) the change in momentum; (b) the average force required.
19. A 1500 kg car skids to a stop from 25.0 m/s in 4.0 s. Find: (a) the impulse on the car; (b) the average braking force.
20. From a force-time graph, a variable force peaks at 1200 N at t = 0.015 s and falls to zero. The area under the graph is estimated at 9.0 N·s. If a 0.20 kg ball was initially at rest, find its final speed.

Conservation of Momentum Problems

21. A 2.0 kg cart moves at +3.0 m/s and collides elastically with a 2.0 kg cart at rest. For equal masses in elastic collision, velocities exchange. Find both final velocities.
22. A 5.0 kg block moves at +4.0 m/s and collides with a 3.0 kg block at −2.0 m/s. They stick together. Find: (a) total initial momentum; (b) final velocity.
23. A 10,000 kg railcar moving at 4.0 m/s collides with and couples to a 6000 kg railcar at rest. Find the final speed.
24. A 75.0 kg astronaut floating in space throws a 1.5 kg toolkit at 8.0 m/s. Find the astronaut’s recoil speed.
25. A 0.30 kg cart and a 0.50 kg cart are held together by a compressed spring on a frictionless track. When released, the 0.30 kg cart moves at 2.0 m/s to the right. Find the velocity of the 0.50 kg cart.
26. A bullet, 0.020 kg, traveling at 400 m/s embeds in a 5.0 kg wooden block at rest. Find: (a) the final velocity of the block+bullet; (b) the kinetic energy before and after; (c) the fraction of kinetic energy lost.

Applied Collision Analysis

27. A forensic reconstruction: Two cars collide at an intersection. Car A, 1200 kg, was heading north; Car B, 1500 kg, was heading east. They lock together and slide 6.0 m at 30° north of east before stopping. If friction decelerated them at 5.0 m/s², find the speeds of both cars just before impact.
28. A 60.0 kg student stands on a 20.0 kg skateboard at rest. She jumps off horizontally at 3.0 m/s to the left relative to the ground. Find the velocity of the skateboard.
29. A proton, with mass 1.67 × 10⁻²⁷ kg, moving at 4.0 × 10⁶ m/s collides head-on with a stationary proton. If this is an elastic collision between equal masses, what are the final velocities?
30. A 3.0 kg ball moving at 5.0 m/s east collides with a 4.0 kg ball moving at 3.0 m/s north. They stick together. Find the magnitude and direction of the combined velocity.

Chapter 16 Checkpoint