Momentum and impulse flashcards

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Momentum
Mass × velocity, mv. A vector: its sign (or direction) matters. Unit N s (= kg m s⁻¹).
Impulse
The change in momentum produced by a force: I = mv − mu. For a constant force acting for time t, I = Ft. Unit N s.
Principle of conservation of momentum
If no external impulse acts on a system of particles, the total momentum of the system is unchanged.
Coalesce
Two particles join together on impact and move on as one particle with a common velocity.
Direct collision
A collision in which both particles move along the same straight line before and after impact.
Impulsive tension
The very large tension acting for a very short time when a slack inextensible string suddenly becomes taut; its impulse gives the particles a common speed along the string.
Impulse–momentum principle
I = mv − mu (u, v signed velocities before and after; one particle). (No — must be known)
Impulse of a constant force
I = Ft. (No — must be known)
Conservation of momentum, two particles in a line
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, all velocities signed with one positive direction. (No — must be known)
Two particles coalesce
m₁u₁ + m₂u₂ = (m₁ + m₂)V. (No — derive it from conservation)
Impulse in vector form
I = mv − mu with I, u, v vectors; work the i and j components separately. (No — must be known)
Why I = Ft for a constant force
F = ma and v = u + at give Ft = m(v − u) = mv − mu. (No — derive it)
Before-and-after diagram for a collision
Two rows (just before, just after): each particle with its mass and a velocity arrow, plus the chosen positive direction.
v–t graph of a particle given an impulse, then slowing by friction
A vertical jump in velocity at the impulse, then a straight line sloping down at gradient −μg to the t-axis.
v–t graph of a pile driver (downwards positive)
Straight line up with gradient g (free fall), sudden drop at impact to the common speed, then a straight line down to zero (driven in against the resistance).
Vector triangle for an impulse
mu followed head to tail by I gives mv: mu + I = mv.
Why is momentum conserved in a collision between two particles?
By Newton's third law they exert equal and opposite forces on each other for the same time, so their impulses are equal and opposite and the total momentum does not change.
Why can you conserve momentum for a collision on a rough surface?
The impact lasts a very short time, so the impulse of friction (or weight) over that time is negligible compared with the collision impulses.
How do the impulses on the two colliding particles compare?
Equal in magnitude, opposite in direction.
A slack string jerks tight over a pulley. Why not add the two momenta?
The particles move in different directions. Write impulse = change in momentum for each particle, with the same impulse J in both.
Particles at rest spring apart. What do you know about their momenta?
The total is still zero, so their momenta are equal in size and opposite in direction.
Where do Newton's law of restitution and loss of kinetic energy in collisions belong?
M2. In M1 you are told enough velocities to use momentum alone.
A 0.3 kg ball hits a wall at 10 m s⁻¹ and rebounds along the same line at 14 m s⁻¹. Magnitude of the impulse?
Away from the wall positive: I = 0.3 × 14 − 0.3 × (−10) = 7.2 N s.
A 0.8 kg particle at rest receives an impulse of 2.4 N s. Speed afterwards?
2.4 = 0.8v, so v = 3 m s⁻¹.
A 5 kg particle at 2 m s⁻¹ coalesces with a 3 kg particle at rest. Common speed?
10 = 8V, V = 1.25 m s⁻¹.
4 kg at 3 m s⁻¹ meets 2 kg at 3 m s⁻¹ head-on. The 4 kg particle stops. Velocity of the 2 kg particle?
12 − 6 = 0 + 2v, so v = 3 m s⁻¹ in the 4 kg particle's original direction (reversed).
A 2.5 kg gun at rest fires a 0.05 kg bullet at 300 m s⁻¹. Recoil speed?
0 = 0.05 × 300 + 2.5w, so the gun recoils at 6 m s⁻¹, opposite to the bullet.
A particle given speed 6 m s⁻¹ slides on rough horizontal ground, μ = 0.2. Time to stop?
Deceleration μg = 1.96 m s⁻²; t = 6 ÷ 1.96 = 3.06 s (or impulse of friction μmg t = 6m).
0.5 kg moving at (2i − j) m s⁻¹ receives an impulse (−1.5i + 2.5j) N s. New velocity and speed?
0.5v = (i − 0.5j) + (−1.5i + 2.5j) = −0.5i + 2j, so v = (−i + 4j) m s⁻¹, speed √17 = 4.12 m s⁻¹.
On a smooth table, A (3 kg) moves at 4 m s⁻¹ directly away from B (2 kg, at rest); the string between them jerks tight. Common speed and impulse in the string?
12 = 5V, V = 2.4 m s⁻¹; impulse on B = 2 × 2.4 = 4.8 N s.
Before writing a momentum equation
Draw before and after, mark the positive direction, sign every velocity. Unknowns go in as positive; the sign of the answer gives the direction.
A particle reverses direction
u and v have opposite signs, so the speeds add in v − u. Using the difference of speeds is the classic slip.
Pile driver: finding the resistance
Include the weight: (M + m)g − R = (M + m)(−a) with downwards positive. Leaving out the weight gives a value too small by (M + m)g.
Multi-stage problem with friction and a collision
Stage 1: friction and constant acceleration to the impact. Stage 2: momentum at the instant of impact. Stage 3: constant acceleration again. Carry unrounded values between stages.
Finding an impulse, which particle should you use?
Either: the impulses are equal and opposite. Use the particle with the simpler numbers and state the direction in words.

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