Newton's Third Law and Conservation of Linear Momentum: Examples and Complete Worked Solutions

Newton’s Third Law and Conservation of Linear Momentum: Examples and Complete Worked Solutions

This companion entry develops the connection between Newton’s Third Law and Conservation of Linear Momentum through worked mechanics problems. The central idea is that internal third-law forces exchange momentum between parts of a system, while the net external force changes the momentum of the system as a whole.

For a system of particles,

∑ P = pi, i

and, when internal forces occur in equal-and-opposite Newtonian pairs,

dP --- = Fext. dt

Therefore an isolated system obeys

P = constant.

Figure 1. A third-law interaction pair acts on two different bodies. The forces cancel only when the two bodies are included in the same system momentum balance.

Part I: Exercises

Exercise 1: identify the third-law pair

A person pushes horizontally on a crate with a force of 120 N to the right. State the corresponding Newton’s-third-law force. On which object does each force act? Explain why the two forces should not both appear on the crate’s Free-body diagram.

Exercise 2: equal forces do not imply equal accelerations

Two carts of masses

m1 = 3.0 kg, m2 = 5.0kg

push on one another with an interaction force of magnitude 20 N. Neglect external horizontal forces. Find the acceleration of each cart and verify that the center of mass has zero horizontal acceleration.

Exercise 3: derive total momentum conservation for two particles

Particles 1 and 2 interact only with each other. Starting from

dp1- dt = F12,

dp2- dt = F21,

and Newton’s third law, derive conservation of total momentum.

Exercise 4: equal and opposite impulse

Two initially stationary carts interact through a constant internal force of magnitude 30 N for 0.40 s. Cart 1 has mass 2.0 kg and cart 2 has mass 3.0 kg. Find the impulse on each cart, the final velocity of each cart, and the final total momentum.

Figure 2. Equal-and-opposite interaction forces produce equal-and-opposite areas under the force-time curves, and therefore equal-and-opposite impulses.

Exercise 5: two skaters push apart

Two skaters initially rest on nearly frictionless ice. Their masses are

m1 = 60kg, m2 = 40 kg.

After they push apart, skater 1 moves at 2.0 m∕s to the right. Find the velocity of skater 2.

Exercise 6: recoil as momentum exchange

A launcher of mass 4.0 kg is initially at rest and ejects a 0.010 kg projectile horizontally at 400 m∕s relative to the ground. Neglect external horizontal impulse during the short launch interval. Find the recoil velocity of the launcher.

Exercise 7: perfectly inelastic collision

A 2.0 kg cart moving at +5.0 m∕s collides with a 3.0 kg cart moving at −1.0 m∕s. The carts stick together. Find their common final velocity. Is kinetic energy conserved?

Exercise 8: elastic collision and momentum exchange

A 2.0 kg cart moving at 6.0 m∕s collides elastically in one dimension with a stationary 3.0 kg cart. Find the final velocities and verify conservation of total momentum.

Figure 3. Collision bookkeeping. The interaction forces act during the collision, but for an isolated two-cart system the total momentum before and after is the same.

Exercise 9: external impulse changes system momentum

A two-cart system has initial total momentum

Pi = 5.0kg m ∕s.

An external horizontal force of 12 N acts on the system for 0.50 s. Find the external impulse and final total momentum.

Exercise 10: choosing the system boundary

A ball strikes a massive wall and rebounds. During the impact the ball’s momentum changes from +3.0 to −2.0 kg m∕s. (a) Is the ball’s momentum conserved? (b) What impulse acts on the ball? (c) Explain how momentum conservation can be restored by enlarging the chosen system boundary.

Exercise 11: many-particle cancellation

Three particles interact pairwise. Write the total internal force sum

F12 + F21 + F13 + F31 + F23 + F32

and use Newton’s third law to show that the net internal force is zero. State the resulting equation for the total momentum when an external force is present.

Exercise 12: center-of-mass acceleration

A system of several particles has total mass

M = 10kg

and net external force

Fext = 20ex − 10ey N.

Find the center-of-mass acceleration.

Exercise 13: two-dimensional explosion

An object initially at rest breaks into three fragments. The fragment data after the explosion are

m = 1.0 kg, v = 3e m ∕s, 1 1 x

m = 2.0 kg, v = 2e m ∕s, 2 2 y

m3 = 1.5 kg.

Find v₃ if external impulse is negligible.

Exercise 14: time-dependent interaction force

Two particles interact through

F12(t) = 4tex N

for 0 ≤ t ≤ 3 s. Find the impulse on particle 1 and the impulse on particle 2. What is the change in the total momentum of the two-particle system?

Exercise 15: when matter momentum alone is not conserved

In an electromagnetic process, the mechanical momentum of matter changes by

ΔPmatter = +0.020ex kg m ∕s.

Suppose no external momentum enters or leaves the complete matter-plus-field system. What change in field momentum is required? Explain why this example shows that momentum conservation is more general than a simple instantaneous mechanical third-law pair.

Exercise 16: synthesis - internal redistribution plus external impulse

Three carts begin at rest. Internal spring forces act while an external impulse

J = 3e N s ext x

is applied to the three-cart system. After all interactions are complete, carts 1 and 2 have momenta

p1 = 4ex kgm ∕s,

p2 = − 2exkg m ∕s.

Cart 3 has mass 3.0 kg. Find p₃ and v₃. Identify which parts of the momentum bookkeeping are internal and which are external.

Figure 4. System boundaries determine which forces count as internal and which count as external. Internal third-law pairs cancel in the system sum; external impulse changes total momentum.

Part II: Complete Worked Solutions

Solution 1: identify the third-law pair

The person exerts on the crate

Fperson on crate = +120ex N.

Newton’s third law gives

F = − 120e N. crate on person x

The first force acts on the crate; the second acts on the person. They are equal and opposite, but they act on different bodies. Therefore only the first belongs on the crate’s free-body diagram.

Solution 2: equal forces do not imply equal accelerations

Take the force on cart 1 to be in the positive x direction:

F12 = +20 N.

Then

F21 = − 20 N.

For cart 1,

20 2 a1 = ---= 6.67m ∕s . 3

For cart 2,

−-20- 2 a2 = 5 = − 4.00m ∕s .

The center-of-mass acceleration is

ACM = m1a1-+-m2a2--. m1 + m2

Thus

A = 3(6.67) +-5-(−-4.00)-= 0. CM 8

Equal interaction forces do not imply equal accelerations; acceleration also depends on mass.

Solution 3: derive total momentum conservation for two particles

Add the two momentum equations:

dp1-+ dp2-= F12 + F21. dt dt

Newton’s third law gives

F + F = 0. 12 21

Therefore

d- dt(p1 + p2) = 0.

Defining

P = p1 + p2,

we obtain

dP --- = 0, dt

P = constant.

Solution 4: equal and opposite impulse

The impulse magnitude is

J = FΔt = (30)(0.40) = 12N s.

Choose the impulse on cart 1 to be positive:

J = +12 N s, 1

J2 = − 12 N s.

The carts start from rest, so

m1v1 = 12.

Hence

12 v1 = ---= 6.0m ∕s. 2

For cart 2,

m v = − 12, 2 2

−-12- v2 = 3 = − 4.0m ∕s.

The final total momentum is

Pf = (2)(6) + (3)(− 4) = 0.

Thus the internal interaction changes the individual momenta but not the total momentum.

Solution 5: two skaters push apart

Initially,

Pi = 0.

With negligible external horizontal impulse,

Pf = 0.

Thus

m v + m v = 0. 1 1 2 2

Substitution gives

(60 )(2) + (40)v = 0. 2

Therefore

v2 = − 3.0m ∕s.

The lighter skater moves faster so that the two momenta are equal in magnitude and opposite in direction.

Solution 6: recoil as momentum exchange

Initial total momentum is zero. Therefore

mpvp + mLvL = 0.

Using

mp = 0.010 kg, vp = 400 m ∕s, mL = 4.0kg,

we obtain

(0.010)(400 ) + (4.0)vL = 0.

Thus

vL = − 1.0 m∕s.

The negative sign means that the launcher recoils opposite the projectile motion.

Solution 7: perfectly inelastic collision

Initial momentum is

Pi = (2)(5) + (3)(− 1) = 7 kg m∕s.

After sticking, the total mass is

M = 5 kg.

Momentum conservation gives

5vf = 7,

vf = 1.40 m ∕s.

The initial kinetic energy is

Ki = 1-(2)(52) + 1-(3)(12) = 26.5 J. 2 2

The final kinetic energy is

K = 1-(5 )(1.42) = 4.90J. f 2

Kinetic energy is not conserved in this perfectly inelastic collision, even though momentum is conserved.

Solution 8: elastic collision and momentum exchange

For a one-dimensional elastic collision with particle 2 initially at rest,

m − m v1f = --1----2v1i, m1 + m2

--2m1---- v2f = m1 + m2 v1i.

Substituting m₁ = 2, m₂ = 3, and v_1i = 6 gives

2-−-3- v1f = 5 (6) = − 1.2 m∕s,

v2f = 4-(6 ) = 4.8 m ∕s. 5

Initial momentum:

P = (2)(6) = 12kg m ∕s. i

Final momentum:

Pf = (2)(− 1.2) + (3)(4.8 ) = − 2.4 + 14.4 = 12kg m ∕s.

Thus total momentum is conserved.

Solution 9: external impulse changes system momentum

The external impulse is

Jext = FextΔt = (12)(0.50 ) = 6.0 N s.

The impulse-momentum theorem for the complete system gives

Pf − Pi = Jext.

Hence

Pf = 5.0 + 6.0 = 11.0 kgm ∕s.

Internal forces do not appear in this total-system equation because their third-law pairs cancel.

Solution 10: choosing the system boundary

For the ball alone,

P = +3.0 kg m ∕s, i

Pf = − 2.0kg m ∕s.

Therefore the ball’s momentum is not conserved.

The impulse on the ball is

J = Pf − Pi = − 2.0 − 3.0 = − 5.0N s.

The wall exerts this external impulse on the ball-only system.

If the chosen system is enlarged to include the ball, wall, and Earth, then the wall-ball interaction is internal. The ball’s momentum loss is accompanied by momentum transferred to the wall/Earth. The total momentum of the enlarged isolated system can remain conserved.

Solution 11: many-particle cancellation

Group the internal forces into third-law pairs:

(F12 + F21) + (F13 + F31) + (F23 + F32).

Each pair vanishes:

Fij + Fji = 0.

Therefore

Finternal,total = 0.

The total momentum equation is consequently

dP --- = Fext. dt

For zero external force, total momentum is constant.

Solution 12: center-of-mass acceleration

The center-of-mass equation is

M A = F . CM ext

Therefore

20ex-−--10ey ACM = 10 .

Hence

ACM = 2ex − 1ey m∕s2.

Internal forces can change the relative motions inside the system, but they do not alter this center-of-mass acceleration when third-law cancellation holds.

Solution 13: two-dimensional explosion

The object starts from rest, so total initial momentum is zero. Therefore

p1 + p2 + p3 = 0.

The first two momenta are

p1 = (1)(3ex) = 3ex,

p2 = (2)(2ey) = 4ey.

Thus

p3 = − 3ex − 4ey kg m ∕s.

Since m₃ = 1.5 kg,

p v3 = -3-. 1.5

Therefore

v3 = − 2.00ex − 2.67ey m∕s.

Solution 14: time-dependent interaction force

The impulse on particle 1 is

∫ 3 J1 = 4t dtex. 0

Thus

[ 2]3 J1 = 2t 0ex = 18exN s.

Newton’s third law gives

F21(t) = − F12(t),

J2 = − 18ex N s.

Therefore

ΔP = J1 + J2 = 0.

Solution 15: when matter momentum alone is not conserved

For the complete matter-plus-field system,

ΔPtotal = ΔPmatter + ΔP field = 0.

Therefore

ΔP field = − 0.020ex kgm ∕s.

Matter momentum alone changed, but total matter-plus-field momentum did not. This is why momentum conservation is more general than the simple Newtonian picture in which every mechanical force on one body is paired instantaneously with an equal-and-opposite mechanical force on another body.

Solution 16: synthesis - internal redistribution plus external impulse

The system initially has zero momentum. The external impulse changes total system momentum by

ΔP = J = 3e kg m ∕s. ext x

Thus the final total momentum must be

Pf = 3ex kgm ∕s.

Using

Pf = p1 + p2 + p3,

we have

3e = 4e − 2e + p . x x x 3

Hence

p3 = 1ex kgm ∕s.

Since m₃ = 3.0 kg,

1- v3 = 3 exm ∕s.

The spring interactions are internal and redistribute momentum among the carts. The 3 N s applied impulse is external and is responsible for the nonzero final momentum of the three-cart system as a whole.

Summary

These examples reinforce four complementary statements:

Newton’s third-law forces act on different bodies.
Internal third-law pairs cancel when the momentum equations of the complete system are added.
The total momentum obeys $dP --- = Fext. dt$
Momentum conservation is the broader organizing principle; in field theories the conserved total may include both matter momentum and field momentum.

References

[1] John R. Taylor, Classical Mechanics, University Science Books, 2005.

[2] Daniel Kleppner and Robert Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.

[3] Stephen T. Thornton and Jerry B. Marion, Classical Dynamics of Particles and Systems, 5th ed., Brooks/Cole, 2004.

[4] Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Vol. I, Basic Books, 2010.

[5] David J. Griffiths, Introduction to Electrodynamics, 4th ed., Pearson, 2013.

"Newton's Third Law and Conservation of Linear Momentum: Examples and Complete Worked Solutions" is owned by bloftin.
(view preamble)