Work, Energy & Momentum — Chapter Practice, Grade 12

90 J

W = Fd = 30 × 3.0 = 90 J.

11760 J

The lifting force equals the weight, mg = 150 × 9.8 = 1,470 N. W = Fd = 1,470 × 8.0 = 11,760 J.

Crane A has greater power

Both cranes do the same work (same load, same height). Since P = W/t, doing that work in less time means Crane A delivers greater power.

3000 W

P = W/t = 18,000 ÷ 6.0 = 3,000 W.

10 m/s

W_net = ΔKE, so 150 = ½ × 3.0 × v² = 1.5v². Then v² = 100 and v = 10 m/s.

State which force's work is calculated.

900/30=30 W.

State what the supplied observations or conditions establish, and separate that from any assumption.

When its component along displacement opposes the displacement, such as kinetic friction on a sliding object.

W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.

Power is energy transferred per time: average P=W/Δt for this process. A watt is one joule per second. Two machines can perform the same work while having different power because one takes less time. A power rating alone does not give total energy without duration and actual operation. Distinguish average power over a stated interval from brief peak power and keep seconds consistent when computing watts.

W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

Useful work=mgh=588 J. Output power=98 W. Input energy=588/0.70=840 J; other transfers/storage total 252 J under the stated boundary.

Useful work=mgh=588 J.

Output power=98 W.

Input energy=588/0.70=840 J; other transfers/storage total 252 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.

100 J

KE = ½mv² = ½ × 2.0 × 10² = ½ × 2.0 × 100 = 100 J.

98 J

PE = mgh = 0.50 × 9.8 × 20 = 98 J.

10 J

PE = ½kx² = ½ × 500 × 0.20² = ½ × 500 × 0.04 = 10 J.

The motorcycle (62,500 J)

Car KE = ½ × 1,000 × 10² = 50,000 J. Motorcycle KE = ½ × 200 × 25² = 62,500 J. The motorcycle's higher speed (squared) wins out over its smaller mass.

Because PE = mgh measures height h from a chosen zero level, the numerical value of PE changes if you pick a different reference. This is not a problem because only changes in PE (as an object rises or falls) affect motion, and those changes come out the same no matter where you place the reference level.

Do not add height to speed or confuse their squares.

½×4×9=18 J.

State what the supplied observations or conditions establish, and separate that from any assumption.

It increases ninefold because of the square dependence.

Us=0.50 J; force magnitude=10 N. Their units and meanings differ.

For an ideal linear spring, elastic potential energy relative to its unstrained state is Us=½kx². This is the area under the balancing force–extension curve, not the endpoint force itself. A spring stretched twice as far stores four times the energy within the same elastic range. Compression can also store energy because x² is positive, while the restoring force reverses direction according to −kx.

Us=0.50 J; force magnitude=10 N. Their units and meanings differ.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

K=½×1.5×16=12 J. Set 12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.

K=½×1.5×16=12 J.

Set 12=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.

120 J

With no friction, mechanical energy is conserved, so all 120 J of PE becomes kinetic energy at the bottom.

10 m/s

All 200 J becomes kinetic energy: 200 = ½ × 4.0 × v² = 2v², so v² = 100 and v = 10 m/s.

70 %

Efficiency = (350 ÷ 500) × 100% = 70%.

Lower than the first hill

Friction dissipates some mechanical energy as heat and sound on the way, so the car reaches the second hill with less energy and cannot climb as high — the second hill must be lower.

As the pendulum swings, energy trades between gravitational potential energy at the high points and kinetic energy at the bottom, with the mechanical total staying nearly constant. A real pendulum stops because friction at the pivot and air resistance gradually dissipate that mechanical energy into thermal energy and sound; the total energy is still conserved, just no longer in the form of visible swinging motion.

Mass cancels in this ideal gravitational-energy calculation.

65 J under the stated accounting.

State what the supplied observations or conditions establish, and separate that from any assumption.

No. Energy or kinematics gives the same speed for the same initial conditions and g.

40 J, because net work is+30 J.

The work–energy theorem states Wnet=ΔK in this ordinary mechanics model. It concerns the sum of all forces’ work on the selected object, not necessarily total energy or one applied force’s contribution. A driving force doing 50 J while resistance does−20 J yields a 30 J kinetic increase. If gravity is among the forces, its work enters this net-work account; a potential-energy formulation reorganizes that term instead of adding it twice.

40 J, because net work is+30 J.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

The accounted final/output total is 190 J, leaving 10 J. Check boundary heat transfer, deformation/storage and instrument uncertainty rather than claiming energy was destroyed.

The accounted final/output total is 190 J, leaving 10 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.

2 kg·m/s

p = mv = 0.50 × 4.0 = 2.0 kg·m/s.

10 kg·m/s

Impulse equals the change in momentum: J = Δp = m(v_f − v_i) = 0.50 × (20 − 0) = 10 kg·m/s.

3 m/s

Conservation of momentum: 3.0 × 4.0 + 1.0 × 0 = (3.0 + 1.0)v′, so 12 = 4.0v′ and v′ = 3.0 m/s.

Momentum only

Momentum is conserved in every collision with no external force. In an inelastic collision, some kinetic energy is converted to heat, sound, and deformation, so only momentum is conserved.

0.6 m/s

Total momentum starts at 0, so 50 × v + 2.0 × 15 = 0, giving 50v = −30 and v = −0.6 m/s. The astronaut recoils at 0.6 m/s in the opposite direction.

Initial KE=9 J and final KE=6 J; sticking need not conserve kinetic energy.

Initial momentum=4; final mass=4; velocity=1 m/s.

State what the supplied observations or conditions establish, and separate that from any assumption.

Momentum is a vector; opposite directions contribute opposite signed components.

vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.

An elastic collision conserves total kinetic energy as well as momentum in the isolated system. An inelastic collision has a kinetic-energy decrease into deformation, thermal changes or other forms; perfectly inelastic means the bodies remain together after impact. “Perfectly” does not mean all kinetic energy disappears. Use momentum first for a shared final velocity, then calculate kinetic energy before and after to determine the change.

vf=ptotal/(mA+mB)=2/3≈+0.667 m/s.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

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.

Write the relationship first; substitute with consistent units.

-15 N·s

3(0−5)=−15.

No; it changes momentum over a longer time and can reduce average force.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

8 N·s

2(1−(−3))=8.

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.

A reversal requires signed velocities, not just a speed difference.

12/0.30=40 N.

State what the supplied observations or conditions establish, and separate that from any assumption.

It extends the time over which the same impulse is delivered; the force–time area remains the momentum change.

Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.

For a fixed momentum change, increasing the stopping interval reduces average net force because Favg=Δp/Δt. The two calculated triangles have equal 3 N·s area but different durations and peaks. Real cushioning can also change rebound, contact geometry and pulse shape, so a longer interval alone does not guarantee a specific peak or injury outcome. Compare actual force histories under controlled conditions rather than treating a simple average-force model as a complete safety prediction.

Rebound changes final velocity and therefore Δp, so the “same momentum change” assumption may no longer hold.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

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.