Momentum flashcards

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Momentum
The product of a body's mass and its velocity, p = mv. A vector, unit kg m s⁻¹.
Principle of conservation of linear momentum
The total momentum of a system of interacting bodies stays constant, provided no resultant external force acts on the system.
Closed (isolated) system
A group of bodies on which no resultant external force acts, so its total momentum is constant.
Internal force
A force that one body in the system exerts on another body in the same system. Internal forces cannot change the total momentum.
External force
A force on the system from something outside it, e.g. friction from the ground or air resistance. A resultant external force changes the total momentum.
Change in momentum
Final momentum − initial momentum: Δp = mv − mu, with velocities substituted with their signs.
Recoil
The motion of one body in the opposite direction when it pushes another body away, e.g. a nucleus emitting a particle or a person throwing an object.
Momentum of a body
p = mv (kg m s⁻¹). (Yes)
Conservation of momentum for two bodies in one dimension
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (velocities with signs). (No — learn it)
Two bodies that stick together
m₁u₁ + m₂u₂ = (m₁ + m₂)v. (No — derive it)
Explosion or push-off from rest
0 = m₁v₁ + m₂v₂, so m₁v₁ = −m₂v₂. (No — derive it)
Newton's second law in terms of momentum
Resultant force = rate of change of momentum, F = Δp/Δt; for constant mass this is F = ma. (No — idea only in Unit 1)
Velocity from a light gate
v = length of card ÷ time for the card to pass through the beam. (No — learn it)
Before-and-after diagram for a collision
Sketch each body before and after with a labelled velocity arrow, and mark the positive direction.
Data-logger v–t graph when trolley A hits and sticks to stationary B
A's line drops and B's line rises during the brief contact; afterwards both lines coincide at the common velocity.
Bar chart of momentum before and after a collision
Individual bars change (equal and opposite amounts) but the total bar is the same height before and after.
Momentum arrows for an explosion from rest
Two arrows of equal length pointing in opposite directions: the total is zero.
How to check a friction-compensated runway with a v–t graph
Give one trolley a push: the v–t graph should be horizontal (constant velocity).
Why does Newton's third law lead to conservation of momentum?
The bodies exert equal and opposite forces on each other for the same time, so their changes in momentum are equal and opposite and the total is unchanged.
Why is momentum a vector?
It is mass (a scalar) × velocity (a vector), so it has the direction of the velocity.
In a collision, which body has the larger change in velocity?
The lighter one: both have the same size of momentum change, which is shared over a smaller mass.
Is kinetic energy conserved whenever momentum is conserved?
No. Momentum is conserved with no resultant external force, but kinetic energy is usually partly transferred to thermal energy, sound or deformation.
Why can friction usually be ignored during a collision?
It is small compared with the forces between the bodies, and the contact time is very short, so it changes the total momentum negligibly.
A ball bounces off a wall. Where does its change of momentum go?
To the wall and the Earth, which gain an equal and opposite momentum; their huge mass makes the velocity change unnoticeable.
Why does a boat move backwards when someone jumps forwards off it?
Total momentum was zero; the person gains forward momentum, so the boat gains equal backward momentum (third-law push on the boat).
Momentum of a 1500 kg car travelling at 22 m s⁻¹?
p = 1500 × 22 = 33 000 kg m s⁻¹ in the direction of travel. (Yes)
A 0.15 kg ball hits a wall at 18 m s⁻¹ and rebounds at 10 m s⁻¹. Size of its change in momentum?
Towards the wall positive: Δp = 0.15 × (−10 − 18) = −4.2 kg m s⁻¹, so 4.2 kg m s⁻¹ away from the wall.
A 2.0 kg trolley at 3.0 m s⁻¹ hits a stationary 1.0 kg trolley and they stick. Common velocity?
2.0 × 3.0 = 3.0v → v = 2.0 m s⁻¹ in the original direction.
A 4.0 kg launcher at rest fires a 0.025 kg ball at 400 m s⁻¹. Recoil velocity of the launcher?
0 = 0.025 × 400 + 4.0v → v = −2.5 m s⁻¹: 2.5 m s⁻¹ opposite to the ball.
3.0 kg at +2.0 m s⁻¹ meets 1.0 kg at −4.0 m s⁻¹ head-on; they stick. Common velocity?
6.0 − 4.0 = 4.0v → v = +0.50 m s⁻¹ (direction of the 3.0 kg body).
0.50 kg glider at 1.2 m s⁻¹ hits a stationary 0.25 kg glider; the first continues at 0.20 m s⁻¹. Speed of the second?
0.60 = 0.10 + 0.25v → v = 2.0 m s⁻¹ in the original direction.
Stating the conservation principle
Two parts: total momentum is constant (before = after) AND no resultant external force acts on the system.
A body moving the other way
Give its velocity a negative sign before you multiply. Adding magnitudes gives the wrong total.
A negative answer for a velocity
Translate it into words: the body moves in the negative direction (e.g. “0.15 m s⁻¹ opposite to the bag”).
Separation of bodies that were already moving
The total momentum before is (m₁ + m₂)u, not zero.
Bodies that move off together
Use the combined mass (m₁ + m₂) after the collision.
Explaining conservation with Newton's laws
Three steps: equal and opposite forces (third law) → same contact time → equal and opposite Δp, total unchanged.
Unit of momentum
kg m s⁻¹, which is equivalent to N s. Do not write kg m s⁻².

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