NCERT Solutions for Class 9th Science Chapter 7 .4.3 Conservation of mechanical energy

Q7.

For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.

Answer

Just before it reaches C the ball has no potential energy left and kinetic energy mgh, so its mechanical energy is still mgh.

The object is released from rest at A, at height h. Take the ground (C) as h = 0.

Height at C: h′ = 0, so potential energy = mg × 0 = 0 J

Speed at C — using v² = u² + 2gh with u = 0:
v² = 0 + 2gh = 2gh

Kinetic energy at C:
K = ½mv² = ½ × m × 2gh = mgh

Mechanical energy at C = K + U = mgh + 0 = mgh ✔

Compare this with the two positions the chapter already worked out:

Point	Potential energy	Kinetic energy	Mechanical energy
A (t = 0, height h)	mgh	0	mgh
B (time t, height h – ½gt²)	mgh – ½mg²t²	½mg²t²	mgh
C (ground, height 0)	0	mgh	mgh

Why it happens: gravity is the only force doing work here. Whatever potential energy the ball loses by falling, it gains back exactly as kinetic energy. The store simply changes its label, so the total never moves off mgh. This is the conservation of mechanical energy.

Check it yourself: take m = 0.5 kg, h = 5 m, g = 10 m s^–2. At A: U = 0.5 × 10 × 5 = 25 J. At C: v = √(2 × 10 × 5) = 10 m s^–1, so K = ½ × 0.5 × 100 = 25 J. Same number.

Q8.

You may have seen an exhibit like that in Fig. 7.22 in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points A, B and C. Why do subsequent points, such as C, D and E, usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?

Answer

The ball is released from rest at the top of the tall tower, so its whole mechanical energy is potential energy there. After that the two forms keep swapping — but the total slowly leaks away.

Each hump is lower than the one before because friction and air resistance remove a little mechanical energy on every stretch of track.

Point	Height	Potential energy	Kinetic energy
A — first hump	high, but below the start	large	small (only the part of the drop already made)
B — valley between humps	lowest, about ground level	almost zero — minimum	maximum — the ball is fastest here
C — next hump	lower than A	large again, but less than at A	more than at A

At every point: K + U = mechanical energy
Going down a slope: U falls, K rises by the same amount
Going up a hump: K falls, U rises by the same amount

Why C, D and E are lower each time — yes, friction is exactly the reason.

Height a ball can climb, h = mechanical energy ÷ mg
On each stretch of track, friction and air resistance do negative work
mechanical energy after = mechanical energy before – (energy lost to friction)
Less mechanical energy → smaller maximum height on the next hump

Why it happens: friction between the ball and the rail, and air resistance, always act opposite to the ball's motion. That negative work converts a little mechanical energy into heat and sound on every pass, and heat and sound cannot climb back into the ball. So the designer must make every hump lower than the last — otherwise the ball would not have enough energy left to get over it and would roll back.

Did you know? Real roller coasters are built the same way. The first drop is always the tallest, and every loop and hill after it is shorter than the one before.

NCERT Solutions for Class 9th Science Chapter 7 .4.3 Conservation of mechanical energy — Pause and Ponder