NCERT Solutions for Class 9th Science Chapter 7 .4.3 Conservation of mechanical energy — Pause and Ponder
Book page 1297 Updated on2026-09-19
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.Fig. 7.19, page 126: an object of mass m dropped from rest at A (height h), passing B (height h′) after time t, and reaching the ground at C.
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?Fig. 7.22, page 129 — redrawn sketch of the ball roller-coaster exhibit: the ball is released at the top of the tall tower, rises to peak A, dips to B, and then over the successively lower peaks C, D and E.
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.