Work, Energy & Momentum — Chapter quiz · Physics & Environmental Science, Grade 12
Account for work, power, stored energy, collisions, and force-time evidence.
Problem 1
Answer and explanation
120 J
W = Fd = 40 × 3.0 = 120 J.
Problem 2
Answer and explanation
1500 W
P = W/t = 6,000 ÷ 4.0 = 1,500 W.
Problem 3
Answer and explanation
36 J
KE = ½mv² = ½ × 2.0 × 6.0² = ½ × 2.0 × 36 = 36 J.
Problem 4
Answer and explanation
117.6 J
PE = mgh = 3.0 × 9.8 × 4.0 = 117.6 J.
Problem 5
- doubles
- halves
- quadruples
- stays the same
Answer and explanation
quadruples
Because KE = ½mv² depends on speed squared, doubling the speed multiplies KE by 2² = 4.
Problem 6
- momentum
- kinetic energy
- potential energy
- temperature
Answer and explanation
kinetic energy
The work-energy theorem is W_net = ΔKE: net work changes an object's kinetic energy.
Problem 7
Answer and explanation
90 J
With no friction, mechanical energy is conserved, so all 90 J of PE becomes kinetic energy at the bottom.
Problem 8
Answer and explanation
75 %
Efficiency = (useful output ÷ input) × 100% = (300 ÷ 400) × 100% = 75%.
Problem 9
- extra kinetic energy
- extra potential energy
- brand-new matter
- thermal energy from friction
Answer and explanation
thermal energy from friction
Friction and drag dissipate mechanical energy mainly as thermal energy (and some sound); the total energy is still conserved.
Problem 10
Answer and explanation
20 kg·m/s
p = mv = 4.0 × 5.0 = 20 kg·m/s.
Problem 11
Answer and explanation
2 m/s
Conservation of momentum: 2.0 × 6.0 + 4.0 × 0 = (2.0 + 4.0)v′, so 12 = 6.0v′ and v′ = 2.0 m/s.
Problem 12
Answer and explanation
In both kinds of collision, momentum is conserved. The difference is kinetic energy: in an elastic collision kinetic energy is also conserved (the objects bounce apart with no energy lost), while in an inelastic collision some kinetic energy is converted into heat, sound, and deformation. In a perfectly inelastic collision the objects stick together and the greatest possible amount of kinetic energy is lost — but momentum is still conserved.
Problem 13
Answer and explanation
W=30 J; P=W/t=15W, if that force is constant. A force perpendicular to displacement does no mechanical work in the point-particle model.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 14
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 15
Answer and explanation
U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences. Gravitational potential energy needs a chosen reference; negative values can be valid.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 16
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 17
Answer and explanation
Energy transfers into thermal energy, deformation, or sound of the chosen larger system; account for boundaries. Mechanical energy can decrease while total energy remains conserved.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 18
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 19
Answer and explanation
Initialp=6 kg·m/s, totalm=3 kg, so vf=2 m/s with negligible external impulse. Momentum conservation does not automatically imply kinetic-energy conservation.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 20
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 21
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
20 N
4/.20=20.
Problem 22
Answer and explanation
Yes, in an inelastic collision 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 23
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.