Impulse and momentum in two dimensions flashcards

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Impulse
The product of a force and the time for which it acts, FΔt. It equals the change of momentum it produces. Vector; unit N s (= kg m s⁻¹).
Newton's second law (momentum form)
The resultant force on a body is equal to its rate of change of momentum, F = Δp/Δt.
Elastic collision
A collision in which the total kinetic energy of the bodies is the same before and after (momentum is also conserved).
Inelastic collision
A collision in which some of the total kinetic energy is transferred to other stores (thermal, sound, deformation); momentum is still conserved.
Glancing (oblique) collision
A collision in which the bodies move off at angles to the original line of motion, so momentum must be conserved in two dimensions.
Principle of conservation of momentum in two dimensions
With no resultant external force, the total momentum is constant, so its component in each of two perpendicular directions is constant.
Impulse–momentum equation
FΔt = Δp (F = average resultant force, Δt = contact time). (Yes)
Kinetic energy in terms of momentum
Ek = p²/2m (non-relativistic speeds). (Yes)
Derivation of Ek = p²/2m
Ek = ½mv² and v = p/m, so Ek = ½m(p/m)² = p²/2m. (No — derive it)
Components of a momentum p at angle θ to the x axis
pₓ = p cos θ, p_y = p sin θ. (No — learn it)
Force from a stream of mass hitting a surface
F = (mass per second) × (change of velocity), from F = Δp/Δt. (No — derive it)
Momentum of a particle of known kinetic energy
p = √(2mEk). (No — rearrange Ek = p²/2m)
What does the area under a force–time graph represent?
The impulse, and so the change of momentum of the body.
Force–time graphs for the same impact with and without padding
Equal areas (same Δp). With padding the pulse is longer and lower: smaller peak and average force.
Momentum vector triangle for a two-body collision (one body at rest)
Momentum before is one side; the two momenta after, drawn head to tail, are the other two sides.
Head-to-tail diagram for an object at rest exploding into three pieces
The three momentum arrows form a closed triangle: the vector sum is zero and they lie in one plane.
Expected graph of impulse (area) against change of momentum in Core Practical 9
A straight line through the origin with gradient 1.
Why does an airbag or crumple zone reduce injury?
The change of momentum is fixed; the padding increases the stopping time, so the average force F = Δp/Δt is smaller.
Why can momentum be conserved separately in x and y?
Momentum is a vector; if the total vector is constant, each of its perpendicular components is constant.
Why is kinetic energy not split into x and y parts?
Kinetic energy is a scalar. Add the kinetic energies of the bodies once, using their speeds.
Two fragments fly apart from rest. Which has more kinetic energy?
The lighter one: momenta are equal in size and Ek = p²/2m, so Ek is inversely proportional to mass.
Equal masses, one at rest, collide elastically. What angle is between their paths afterwards?
90°: momentum gives u = v_A + v_B (vectors) and kinetic energy gives u² = v_A² + v_B², which is Pythagoras.
Is momentum lost in an inelastic collision?
No. Momentum is conserved whenever there is no resultant external force; only kinetic energy is transferred to other stores.
In Core Practical 10, why must the scale be in the same plane as the spheres?
So distances on the video are calibrated correctly, with no parallax error from a scale nearer to or further from the camera.
In Core Practical 9, why is the runway tilted slightly?
So a component of weight balances friction: there is no resultant external force before the impact, so the light-gate speed is the impact speed.
A 0.15 kg ball hits a bat at 12 m s⁻¹ and leaves along the same line at 8.0 m s⁻¹ the other way. Contact 0.010 s. Average force?
Δp = 0.15 × (8.0 + 12) = 3.0 N s; F = 3.0 ÷ 0.010 = 300 N. (Yes)
A triangular force–time pulse peaks at 80 N and lasts 0.040 s. It acts on a 0.40 kg body at rest. Final speed?
Impulse = ½ × 80 × 0.040 = 1.6 N s; v = 1.6 ÷ 0.40 = 4.0 m s⁻¹.
0.30 kg at 4.0 m s⁻¹ east and 0.40 kg at 3.0 m s⁻¹ north collide and stick. Common speed?
East 1.2, north 1.2 kg m s⁻¹; total √(1.2² + 1.2²) = 1.70 kg m s⁻¹; v = 1.70 ÷ 0.70 = 2.4 m s⁻¹ at 45° north of east.
Kinetic energy of a proton (1.67 × 10⁻²⁷ kg) with momentum 4.0 × 10⁻²⁰ kg m s⁻¹?
Ek = p²/2m = (4.0 × 10⁻²⁰)² ÷ (2 × 1.67 × 10⁻²⁷) = 4.8 × 10⁻¹³ J. (Yes)
2.0 kg at 3.0 m s⁻¹ sticks to a stationary 1.0 kg body. Kinetic energy before and after?
v = 6.0 ÷ 3.0 = 2.0 m s⁻¹. Before 9.0 J, after ½ × 3.0 × 2.0² = 6.0 J: inelastic, 3.0 J transferred.
A 0.40 kg object at rest splits: 0.10 kg at 2.0 m s⁻¹ along +x, 0.10 kg at 2.0 m s⁻¹ along +y. Speed of the 0.20 kg piece?
Its momentum is √(0.20² + 0.20²) = 0.283 kg m s⁻¹, so v = 0.283 ÷ 0.20 = 1.4 m s⁻¹, at 45° between −x and −y.
Video at 100 frames per second: a sphere moves 30 mm in 5 frame intervals. Speed?
Time = 5 ÷ 100 = 0.050 s; v = 0.030 ÷ 0.050 = 0.60 m s⁻¹.
Two-dimensional collision method
Choose x and y axes, resolve every momentum, write one conservation equation for each axis, then recombine with Pythagoras and tan⁻¹.
Force–time graph with time in ms
Convert to seconds before finding the area, or the impulse is 1000 times too big.
“Show that the collision is inelastic”
Calculate total kinetic energy before and after and compare the numbers; a sentence alone is not enough.
A ball bouncing off a smooth wall at an angle
Only the component perpendicular to the wall reverses: Δp = 2mv cos θ, with θ measured from the normal.
Combining uncertainties in p = mL/t (Unit 4)
Add the percentage uncertainties in m, L and t.
Giving a direction in two dimensions
State an angle and the line it is measured from, e.g. “29.9° to A's original direction, on the other side from A”.

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