2.10-2.12
Your guide: Sir HamzaBelieves every formula has a story, and units never lie.
A car crash, a karate chop, a rocket launch. All use momentum.
Let us finish the chapter with these three.
Smart Syllabus 2026 leaves out sections 2.10 to 2.12 for the 2026 exam. Read them for understanding, not for marks.
In real collisions the bodies often fly off at angles. Momentum is a vector. So we split it into x and y parts.
Kinetic energy is a scalar. It needs no split.
After the hit, m1 and m2 go at angles θ1 and θ2 to the x axis.
Along x: m1v1 + m2v2 = m1v1′ cos θ1 + m2v2′ cos θ2 (Eq. 2.35).
Along y: 0 = m1v1′ sin θ1 − m2v2′ sin θ2 (Eq. 2.36).
Energy: ½m1v12 + ½m2v22 = ½m1(v1′)2 + ½m2(v2′)2 (Eq. 2.37).
Big collisions in daily life are mostly inelastic. In the book's perfect inelastic case the bodies stick. Mass M = m1 + m2 moves at v_f, at angle φ to the x axis.
x: m1v1 + m2v2 cos θ = Mv_f cos φ (Eq. 2.38).
y: 0 + m2v2 sin θ = Mv_f sin φ (Eq. 2.39).
Kinetic energy before: (K.E.)ᵢ = ½m1v12 + ½m2v22 (Eq. 2.40). After: (K.E.)_f = ½Mv_f2 (Eq. 2.41).
The loss is ΔK.E. = (K.E.)ᵢ − (K.E.)_f. It goes to heat, sound or change of shape.
The book gives three everyday cases: a karate chop on bricks, a car crash, and a ball hit by a bat.
A rocket throws hot gas backward at high speed. The gas gains momentum backward. The rocket gains equal momentum forward.
Let m be the mass of gas thrown per second, and v its speed relative to the rocket. The rocket's acceleration is a = mv ÷ M (Eq. 2.42). M is the rocket's mass.
As fuel burns, M falls. So the acceleration rises.
The book says a typical rocket burns about 10 000 kg of fuel each second, at over 4000 m s−1. More than 80% of the launch mass is fuel. Multi-stage rockets drop empty stages.
Step 1 / 7
A car crash, a karate chop, a rocket launch. All use momentum.
Let us finish the chapter with these three.
Smart Syllabus 2026 leaves out sections 2.10 to 2.12 for the 2026 exam. Read them for understanding, not for marks.
In real collisions the bodies often fly off at angles. Momentum is a vector. So we split it into x and y parts.
Kinetic energy is a scalar. It needs no split.
After the hit, m1 and m2 go at angles θ1 and θ2 to the x axis.
Along x: m1v1 + m2v2 = m1v1′ cos θ1 + m2v2′ cos θ2 (Eq. 2.35).
Along y: 0 = m1v1′ sin θ1 − m2v2′ sin θ2 (Eq. 2.36).
Energy: ½m1v12 + ½m2v22 = ½m1(v1′)2 + ½m2(v2′)2 (Eq. 2.37).
Big collisions in daily life are mostly inelastic. In the book's perfect inelastic case the bodies stick. Mass M = m1 + m2 moves at v_f, at angle φ to the x axis.
x: m1v1 + m2v2 cos θ = Mv_f cos φ (Eq. 2.38).
y: 0 + m2v2 sin θ = Mv_f sin φ (Eq. 2.39).
Kinetic energy before: (K.E.)ᵢ = ½m1v12 + ½m2v22 (Eq. 2.40). After: (K.E.)_f = ½Mv_f2 (Eq. 2.41).
The loss is ΔK.E. = (K.E.)ᵢ − (K.E.)_f. It goes to heat, sound or change of shape.
The book gives three everyday cases: a karate chop on bricks, a car crash, and a ball hit by a bat.
A rocket throws hot gas backward at high speed. The gas gains momentum backward. The rocket gains equal momentum forward.
Let m be the mass of gas thrown per second, and v its speed relative to the rocket. The rocket's acceleration is a = mv ÷ M (Eq. 2.42). M is the rocket's mass.
As fuel burns, M falls. So the acceleration rises.
The book says a typical rocket burns about 10 000 kg of fuel each second, at over 4000 m s−1. More than 80% of the launch mass is fuel. Multi-stage rockets drop empty stages.
Q1. Why does a rocket's acceleration rise as it burns fuel?
a = mv ÷ M. M falls, so a rises.
✗ “Kinetic energy is conserved in every collision.”
✓ Only in elastic collisions. In inelastic ones some is lost.
✗ “Split kinetic energy into components too.”
✓ Kinetic energy is a scalar. Only momentum is split.
1. In 2-D collisions, momentum is conserved:
(c) Eq. 2.35 and 2.36 give both.
2. Rocket acceleration is:
(a) Thrust mv divided by the rocket's mass M.
3. Two bodies stick together after the hit. This is:
(b) The book calls this a perfect inelastic collision.
4. The kinetic energy lost is:
(a) Before minus after, as in the book.
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