Linear momentum is a vector. Choose a positive direction first, attach a sign to velocity, and the momentum sign follows automatically. A complete answer includes magnitude, direction, and units.
Classical momentump = mv
Momentum is mass times velocity — how much 'moving matter' there is, and in which direction.
Mass in kg, velocity in m/s
Fixed-mass changeΔp = m(v − u)
How much the momentum changed: mass times the difference between final speed (v) and starting speed (u).
Final momentum minus initial momentum
Equivalent unit1 kg·m/s = 1 N·s
Momentum's unit equals force × time — a hint that forces acting over time are what change momentum.
Momentum and impulse share equivalent units
01
Direction matters
Equal speeds in opposite directions give equal momentum magnitudes but opposite momentum vectors.
02
Choose a frame
Velocity—and therefore momentum—depends on the observer's reference frame. State the frame when it matters.
03
Add vectors
System momentum is the vector sum of every object's momentum, not the sum of their magnitudes.
02
Change one variable at a time
Make the vector account visible.
Use a negative velocity to reverse the cart. Zero speed gives zero momentum even when the mass is large.
0
2.0 kg
Momentum vector
+10 kg·m/s
p = mv+10 kg·m/s
K = 1/2 mv²25 J
03
Catch the common trap
Explain before calculating.
Two identical carts move at equal speeds in opposite directions. How do their momenta compare?
Choose an answer, then use the explanation to check your model.
04
Worked examples
Choose signs, balance, then check.
EasyFind the momentum of a 1500 kg car travelling at 20 m/s.
p = mv = 1500 × 20.
p = 3.0 × 10⁴ kg·m/s in the direction of motion.
Answerp = 3.0 × 10⁴ kg·m/s
MediumA 0.20 kg ball changes velocity from +12 m/s to −8.0 m/s. Find its change in momentum.
Choose the original direction as positive.
pi = (0.20)(+12) = +2.4 kg·m/s and pf = (0.20)(−8.0) = −1.6 kg·m/s.
Δp = pf − pi = −1.6 − 2.4.
AnswerΔp = −4.0 kg·m/s, or 4.0 kg·m/s opposite the original motion
HardA 60 g tennis ball at 40 m/s and a 7.0 kg bowling ball share the same momentum. Find the bowling ball's speed and compare their kinetic energies.
p = 0.060 × 40 = 2.4 kg·m/s, so v = 2.4 ÷ 7.0 ≈ 0.343 m/s.
K(tennis) = ½ × 0.060 × 1600 = 48 J.
K(bowling) = ½ × 7.0 × 0.1176 ≈ 0.41 J — same momentum, wildly different energies.
Answerv ≈ 0.34 m/s; the tennis ball carries ~117× the kinetic energy
ChallengingRain falls vertically into an open truck rolling freely at 8.0 m/s. After collecting 200 kg of water, the 800 kg truck's speed has changed. Find it, and explain what happened to the 'lost' motion.
Horizontal momentum is conserved: 800 × 8.0 = (800 + 200) × v.
v = 6400 ÷ 1000 = 6.4 m/s.
The rain had no horizontal momentum; sharing the same total among more mass slows the truck, and the accompanying kinetic-energy drop becomes heat in the splashes.
Answerv = 6.4 m/s — momentum conserved, kinetic energy not
Exam diagrams for this topic1 figure to inspect and practiseQuestions, hints and marking points in one compact subsection.Open subsectionHide subsection
See it. Read it. Work it.
These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.
Fig. 8.1Two panels, one above the other. In the upper panel, labelled before the collision, trolley A of mass 0.80 kg stands on a level track with an arrow showing it moving to the right at 2.5 m/s towards trolley B of mass 1.2 kg, which is at rest. In the lower panel, labelled after the collision, the two trolleys are drawn joined together and moving to the right with a speed marked v.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. In the upper panel the 2.5 m/s belongs to A alone, B at rest carrying none; in the lower panel v is the speed of the joined 2.0 kg, not of either trolley on its own.
aState State the momentum of trolley B before the collision, and give the reason from the upper panel of Fig. 8.1.
recall2 marks
Check answer · 2 marks
zero (0 kg m/s) (1)
B is at rest, so its velocity is zero, and momentum is mass x velocity (1)
bDetermine After the collision the joined trolleys move to the right at 1.0 m/s. Determine the change in momentum of trolley A, and give its direction.
routine3 marks
Check answer · 3 marks
momentum of A before = 0.80 × 2.5 = 2.0 kg m/s to the right (1)
momentum of A after = 0.80 × 1.0 = 0.80 kg m/s to the right (1)
change = 1.2 kg m/s, directed to the left, that is opposite to A's motion (1)
cCalculate The two trolleys are in contact for 0.15 s. Calculate the average force that trolley B exerts on trolley A during the collision.
demanding3 marks
Check answer · 3 marks
force = change in momentum / time taken (1)
= 1.2 / 0.15 (1)
= 8.0 N, acting to the left on A (1)
dExplain Using both panels of Fig. 8.1, explain why the momentum gained by trolley B is exactly equal to the momentum lost by trolley A, even though B is the heavier trolley and its velocity changes by less.
top of the paper4 marks
Check answer · 4 marks
the force B exerts on A is equal in size and opposite in direction to the force A exerts on B (1)
the two trolleys are in contact for the same length of time, so force x time is the same for both and the changes in momentum are equal and opposite (1)
A's velocity changes by 1.5 m/s, giving 0.80 × 1.5 = 1.2 kg m/s; B's changes by 1.0 m/s, giving 1.2 × 1.0 = 1.2 kg m/s (1)
B's larger mass is offset exactly by its smaller change in velocity, so the total momentum of the two trolleys is unchanged (1)
Transfer challenge
A skater of mass 50 kg stands at rest on ice holding a ball of mass 2.0 kg. She throws the ball horizontally away from her at 6.0 m/s. Determine the speed at which she moves backwards, and state the total momentum of the skater and ball after the throw.
Check answer · 4 marks
momentum of ball after the throw = 2.0 × 6.0 = 12 kg m/s (1)
the skater must carry 12 kg m/s in the opposite direction, since the total was zero before (1)
speed = 12/50 = 0.24 m/s (1)
total momentum after the throw = zero, the same as before (1)