Work, Energy & Momentum — Chapter review · Physics & Environmental Science, Grade 12
Account for work, power, stored energy, collisions, and force-time evidence.
Review the key ideas
Work & Power
Work is energy transferred by a force acting through a displacement (W = Fd·cosθ); the net work on an object changes its kinetic energy; power (P = W/t) is how fast that work is done.
Kinetic & Potential Energy
Kinetic energy depends on speed/frame; potential energy depends on interaction and chosen reference. Conserve a consistent energy account rather than mixing force with energy.
Conservation of Energy
Total energy follows a complete boundary account. Mechanical energy remains constant only under appropriate transfer conditions; useful efficiency needs a defined service.
Momentum & Collisions
System momentum change equals external impulse. Negligible external impulse allows momentum conservation; kinetic energy is also conserved only in the elastic collision model.
Impulse & Collision Evidence
Impulse is signed momentum change; the relevant system boundary determines conservation.
Problem 1
Answer and explanation
200 J
W = Fd = 25 × 8.0 = 200 J.
Problem 2
Answer and explanation
Work = 2,100 J; power = 420 W
Work against gravity: W = Fd = 700 × 3.0 = 2,100 J. Power: P = W/t = 2,100 ÷ 5.0 = 420 W.
Problem 3
Answer and explanation
400 N
P = Fv, so F = P/v = 12,000 ÷ 30 = 400 N. (At constant speed the driving force equals the resistive force.)
Problem 4
Answer and explanation
40 J
KE = ½mv² = ½ × 5.0 × 4.0² = ½ × 5.0 × 16 = 40 J.
Problem 5
Answer and explanation
1.5 J
PE = ½kx² = ½ × 300 × 0.10² = ½ × 300 × 0.01 = 1.5 J.
Problem 6
Answer and explanation
PE = 196,000 J; speed = 28 m/s
PE = mgh = 500 × 9.8 × 40 = 196,000 J. All of it becomes kinetic energy: 196,000 = ½ × 500 × v² = 250v², so v² = 784 and v = 28 m/s.
Problem 7
Answer and explanation
75% efficient; 300 J lost
Efficiency = (900 ÷ 1,200) × 100% = 75%. Energy lost = 1,200 − 900 = 300 J, dissipated as heat.
Problem 8
Answer and explanation
10 m/s
W_net = ΔKE, so 250 = ½ × 5.0 × v² = 2.5v². Then v² = 100 and v = 10 m/s.
Problem 9
Answer and explanation
Useful work = 29,400 J; efficiency = 73.5%
Useful work = gain in gravitational PE = mgh = 200 × 9.8 × 15 = 29,400 J. Efficiency = (29,400 ÷ 40,000) × 100% = 73.5%.
Problem 10
Answer and explanation
16 kg·m/s
p = mv = 2.0 × 8.0 = 16 kg·m/s.
Problem 11
Answer and explanation
The incoming ball stops (0 m/s); the struck ball moves off at 6.0 m/s.
For a head-on elastic collision between equal masses, the objects exchange velocities. Check: momentum 1.0 × 6.0 = 1.0 × 6.0 ✓, and kinetic energy ½ × 1.0 × 6.0² = ½ × 1.0 × 6.0² ✓. This is exactly the behavior seen in a Newton's cradle.
Problem 12
Answer and explanation
2.6 m/s
Conservation of momentum: 2.0 × 5.0 + 3.0 × 1.0 = (2.0 + 3.0)v′, so 10 + 3 = 5.0v′, giving 13 = 5.0v′ and v′ = 2.6 m/s.
Problem 13
Answer and explanation
2.0 kg·m/s
Taking the outgoing direction as positive, Δp = m(v_f − v_i) = 0.20 × (5.0 − (−5.0)) = 0.20 × 10 = 2.0 kg·m/s. (Bouncing back doubles the impulse compared with simply stopping.)
Problem 14
Answer and explanation
W=30 J; P=W/t=15W, if that force is constant.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 15
Answer and explanation
Useful work=mgh=588 J. Output power=98 W. Input energy=588/0.70=840 J; other transfers/storage total252 J under the stated boundary.
Useful work=mgh=588 J.
Output power=98 W.
Input energy=588/0.70=840 J; other transfers/storage total252 J under the stated boundary.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 16
Answer and explanation
U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 17
Answer and explanation
K=½×1.5×16=12 J. Set12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.
K=½×1.5×16=12 J.
Set12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 18
Answer and explanation
Energy transfers into thermal energy, deformation, or sound of the chosen larger system; account for boundaries.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 19
Answer and explanation
The accounted final/output total is190 J, leaving10 J. Check boundary heat transfer, deformation/storage and instrument uncertainty rather than claiming energy was destroyed.
The accounted final/output total is190 J, leaving10 J.
Check boundary heat transfer, deformation/storage and instrument uncertainty rather than claiming energy was destroyed.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 20
Answer and explanation
Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 21
Answer and explanation
Initial momentum=12 kg·m/s, so vf=3 m/s. Ki=24 J and Kf=18 J;6 J leaves the kinetic category through other energy forms.
Initial momentum=12 kg·m/s, so vf=3 m/s.
Ki=24 J and Kf=18 J;6 J leaves the kinetic category through other energy forms.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 22
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
1 N·s
.5×.020×100=1.
Problem 23
Answer and explanation
External forces can transfer momentum into or out of the selected system.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 24
Answer and explanation
Stopping impulse=0.50(0−6)=−3 N·s and Favg=−20 N. With rebound, impulse=0.50(−2−6)=−4 N·s, so the fixed-impulse comparison no longer applies.
Stopping impulse=0.50(0−6)=−3 N·s and Favg=−20 N.
With rebound, impulse=0.50(−2−6)=−4 N·s, so the fixed-impulse comparison no longer applies.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.