3.1 Translational Kinetic Energy — Worked Solutions
Work through Topic 3.1
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A cart speed doubles while mass stays fixed. What happens to K?
- It doubles
- It quadruples
- It stays the same
Step 1. Write K_old = ½mv² and K_new = ½m(2v)².
Why this step? Mass is fixed, so the only changed factor in translational kinetic energy is the squared speed.
Step 2. The ratio is K_new/K_old = 4v²/v² = 4.
Why this step? Squaring the factor of two quadruples energy; a linear factor of two would ignore the speed-squared dependence.
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Question 2
A cart reverses velocity from +3 to −3 m/s at the same speed. Compare its endpoint kinetic energies.
- The final energy is negative
- The energies are equal
- The final energy is three times larger
Step 1. At the two endpoints, v² is (+3)² = 9 and (−3)² = 9 in m²/s².
Why this step? The question compares endpoint energies at equal speed magnitude; the direction signs disappear on squaring.
Step 2. With the same mass, both endpoints have K = ½m·9, so the energies are equal.
Why this step? A scalar kinetic energy does not take the velocity’s sign; this says nothing about possible energy changes during the reversal itself.
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Question 3
A moving observer measures a different cart speed. How should you find its kinetic energy in that frame?
- Transform velocity, then use ½mv²
- Subtract the observer’s kinetic energy from the ground value
- Keep the ground kinetic energy in every frame
Step 1. In one dimension, first find v_relative = v_cart − v_observer using the same axis.
Why this step? K describes motion measured by that observer, so the velocity in its defining equation must belong to that observer’s frame.
Step 2. Then calculate K_relative = ½m(v_relative)².
Why this step? The entire velocity difference is squared; a velocity offset cannot be applied as a simple kinetic-energy subtraction.
Step 3. For example, v_cart = 6 m/s and v_observer = 4 m/s give 2 J for a 1 kg cart, not 18 − 8 = 10 J.
Why this step? The mismatch exposes the missing cross term in subtracting separate squares, so it rejects the proposed energy-subtraction shortcut.
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Question 4
A 3 kg cart moves at 4 m/s. Find K. What speed would a 12 kg cart need to have the same kinetic energy?
Step 1. For the 3 kg cart, K = ½(3)(4²) = 24 J.
Why this step? The squared 4 m/s speed and 3 kg mass give the translational motion energy, with units kg·m²/s².
Step 2. Set the second energy equal: 24 = ½(12)v_B², giving v_B² = 4 m²/s².
Why this step? Equal energy constrains the product m_Bv_B²; the heavier mass must have a smaller squared speed.
Step 3. Take the speed magnitude v_B = 2 m/s.
Why this step? The positive square root reports speed, while the equation itself cannot select a direction of travel.
Step 4. Check: four times the mass and half the speed give factor 4·(½)² = 1 in K.
Why this step? This scaling reproduces equal energy and guards against the incorrect direct-inverse-speed answer of 1 m/s.
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Question 5
Sketch K versus signed velocity for one mass, marking velocities −2, 0, and +2 m/s. Explain the symmetry and why K isn’t a vector.
Step 1. Plot K = ½mvₓ² for the fixed positive mass: an upward parabola with minimum at vₓ = 0.
Why this step? A positive coefficient times a squared component is nonnegative and reaches its smallest value when the component vanishes.
Step 2. Mark equal heights ½m(2 m/s)² at vₓ = −2 m/s and +2 m/s, and zero at vₓ = 0.
Why this step? The opposite signed components have equal squares, so mirror points about zero must have the same energy.
Step 3. The symmetry shows that K has no leftward or rightward direction.
Why this step? A directional vector component would reverse sign under reversal; kinetic energy instead retains the same scalar value.
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Question 6
A 1 kg cart moves +6 m/s relative to ground. Find its kinetic energy measured from a frame moving +4 m/s, and explain why subtracting ½(1)(4²) from its ground energy would fail.
Step 1. Transform velocity: v_relative = 6 − 4 = +2 m/s.
Why this step? The observer moves in the same positive direction, so its motion removes 4 m/s from the cart’s ground-relative component.
Step 2. Compute K_relative = ½(1)(2²) = 2 J.
Why this step? Use the transformed velocity in the kinetic-energy definition; the mass is unchanged in this ordinary nonrelativistic calculation.
Step 3. The ground value is 18 J; subtracting ½(1)(4²) = 8 J would give 10 J, which is wrong.
Why this step? The correct square is (6 − 4)² = 36 − 48 + 16, not 36 − 16; energy subtraction misses the velocity cross term.