Momentum and Impulse
Track direction, connect impulse to momentum change, and conserve total momentum across collisions and recoil.
Momentum and Impulse
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Momentum depends on mass and velocity; impulse changes it
Momentum and impulse are vectors, so direction matters. Choose one direction as positive before substituting values. A rebound produces a larger momentum change than merely stopping because the final velocity has the opposite sign.
Total momentum is conserved when the system receives no significant external impulse. Kinetic energy, however, is not necessarily conserved. When objects stick together, some kinetic energy becomes heat, sound, or deformation.
Impulse is an area and a change
For constant force, impulse is FΔt. On a force–time graph, it is the signed area under the graph. Either way, it equals Δp.
Longer stopping time lowers force
Seat belts, airbags, helmets, padded mats, and follow-through increase the time for a required momentum change, reducing average force.
Write a sign on every velocity before using momentum conservation
Problem: A 2 kg cart moving at +5 m/s sticks to a 3 kg stationary cart.
Before:
p = 2(+5) + 3(0) = +10 kg·m/sAfter: The combined mass is 5 kg.
+10 = 5v → v = +2 m/sAnswer: The joined carts move at 2 m/s in the original direction.
Why it works
The sign records direction, and combining the masses only after a sticking collision prevents the most common setup errors.
Five forms you should recognize
Problem: A 2 kg cart moves at 6 m/s; a 4 kg cart moves at 3 m/s in the same direction.
p₁ = 2(6) = 12 kg·m/sp₂ = 4(3) = 12 kg·m/sDifferent masses and speeds can produce equal momentum.
Problem: A constant 15 N force acts for 0.20 s.
J = FΔt = 15(0.20) = 3 N·sThe object’s momentum changes by 3 kg·m/s in the force direction.
Problem: A 0.50 kg ball changes from +8 m/s to −6 m/s.
Δp = 0.50(−6 − 8) = −7 kg·m/sThe negative sign shows that the impulse points opposite the ball’s original motion.
Problem: The passenger undergoes the same change from moving to rest with or without an airbag.
The airbag increases Δt. Since Favg = Δp/Δt, the average force decreases.
Problem: A stationary system separates into two moving parts.
Its initial momentum was zero. Therefore, one part’s forward momentum must be balanced by equal backward momentum of the other part.
Check before you commit
- Ignoring direction when subtracting velocities
- Using speed instead of signed velocity in a rebound
- Assuming momentum and kinetic energy are both conserved in every collision
- Combining masses before confirming that objects stick
- Saying airbags reduce the required impulse to stop the passenger
- Forgetting that momentum conservation applies to a defined system with negligible external impulse
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Momentum and Impulse FAQ
Are N·s and kg·m/s equivalent?
Yes. Because 1 N = 1 kg·m/s², multiplying by seconds gives kg·m/s.
Is momentum conserved when objects stick?
Yes, for an isolated system. Kinetic energy is not fully conserved in a sticking collision.
Why is a rebound’s momentum change large?
The velocity reverses sign, so the final and initial momenta subtract across opposite directions.
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