Conservation of Mechanical Energy — IISER Physics MCQs with Solutions
Free IISER Physics Conservation of Mechanical Energy MCQs with step-by-step solutions (8 questions). Part of Work, Power and Energy. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Conservation of Mechanical Energy · easy · theory
A mass oscillating on a spring (no friction) has its maximum kinetic energy:
A. At the mean (equilibrium) position ✓ Correct
B. At the extreme positions
C. The KE is constant
D. Halfway to the extreme
Solution: KE + ½kx² = constant; KE is largest where the spring PE is least — at x = 0 (the mean position).
Q2 — Conservation of Mechanical Energy · medium · theory
A roller-coaster car climbs a hill with the motor OFF (smooth track). As it rises, its:
A. Speed stays constant
B. Speed decreases as KE converts to PE ✓ Correct
C. PE decreases
D. Total mechanical energy decreases
Solution: With no engine or friction, ½mv² + mgh = constant, so rising (h↑) costs speed (v↓).
Q3 — Conservation of Mechanical Energy · easy · numerical
A 2 kg block moving at 10 m/s on a smooth floor hits a spring of k = 800 N/m. The maximum compression of the spring is:
A. 2 m
B. 1 m
C. 0.25 m
D. 0.5 m ✓ Correct
Solution: ½mv² = ½kx² ⇒ x = v√(m/k) = 10√(2/800) = 10×0.05 = 0.5 m.
Q4 — Conservation of Mechanical Energy · medium · numerical
A 0.5 kg ball is projected vertically up with 40 J of kinetic energy. The maximum height reached is:
A. 8 m ✓ Correct
B. 16 m
C. 4 m
D. 80 m
Solution: KE → PE: h = KE/mg = 40/(0.5×10) = 8 m.
Q5 — Conservation of Mechanical Energy · hard · numerical
A 1 kg bob on a string is released from the horizontal position. At the lowest point of the swing, the tension in the string is:
A. 30 N ✓ Correct
B. 20 N
C. 40 N
D. 10 N
Solution: v² = 2gL at the bottom; T − mg = mv²/L = 2mg ⇒ T = 3mg = 30 N. Trap: T ≠ mg — the bob is accelerating centripetally.
Q6 — Conservation of Mechanical Energy · medium · numerical
A projectile is fired at 20 m/s. At the top of its path its speed is 12 m/s (the horizontal component). The maximum height reached is:
A. 7.2 m
B. 20 m
C. 12.8 m ✓ Correct
D. 16 m
Solution: Energy per kg: h = (u² − v_top²)/2g = (400 − 144)/20 = 12.8 m.
Q7 — Conservation of Mechanical Energy · hard · theory
A ball is thrown up WITH air resistance present. Compared with the ideal case:
A. It rises higher
B. It returns with the same speed
C. It rises less high, and returns with a smaller speed than it was thrown ✓ Correct
D. Time up exceeds time down
Solution: Air drag removes mechanical energy on both legs: lower peak, slower return; ascent is quicker than descent (larger deceleration going up).
Q8 — Conservation of Mechanical Energy · hard · numerical
A pendulum of length 1.25 m is released from 60° to the vertical. The bob's speed at the lowest point is (cos60° = 0.5):
A. ≈ 4.3 m/s
B. 5 m/s
C. 2.5 m/s
D. ≈ 3.5 m/s ✓ Correct
Solution: h = L(1 − cos60°) = 1.25×0.5 = 0.625 m; v = √(2×10×0.625) = √12.5 ≈ 3.5 m/s.