A ball of mass 100 g dropped on the ground from a height of 5 m bounces repeatedly. During every bounce, 64% o

Numericals · Q40

Q. A ball of mass 100 g dropped on the ground from a height of 5 m bounces repeatedly. During every bounce, 64% of the potential energy (just before that bounce) is converted into kinetic energy (just after that bounce). Calculate the following:

(a) Coefficient of restitution.

(b) Speed with which the ball comes up from the ground after the third bounce.

(d) Average force exerted by the ground if this impact lasts for 250 ms.

(e) Average pressure exerted by the ball on the ground during this impact if the contact area of the ball is 0.5 cm^2.

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Concept understanding — Coefficient of Restitution

The Bounce: What Makes a Ball Bounce Back?

Drop a rubber ball and it springs back up. Drop a lump of clay and it just splats. The difference between these two extremes is captured by a single number: the coefficient of restitution (often written as $e$ ).

The intuition is simple. When two objects collide, they always deform a little — like a spring getting squashed. The question is: how much of that "squash" energy gets given back as motion? A perfect spring gives everything back; a lump of clay gives nothing back. The coefficient of restitution measures exactly that.

The Precise Definition

For any collision along a straight line (what we call head-on or one-dimensional collision), the coefficient of restitution is defined as:

e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}

Let's unpack that. Suppose two balls, A and B, move toward each other. Before the collision, their velocities are $u_A$ and $u_B$ . After the collision, they are $v_A$ and $v_B$ .

The relative velocity of approach is how fast they are coming together: $u_A - u_B$ (taking direction into account — if they move toward each other, this is positive).

The relative velocity of separation is how fast they are moving apart after the collision: $v_B - v_A$ .

So the formula becomes:

e = \frac{v_B - v_A}{u_A - u_B}

Important

The coefficient of restitution is a dimensionless number (no units) that always lies between 0 and 1 for real collisions.

What the Numbers Mean

$e = 1$ — Perfectly Elastic Collision

No kinetic energy is lost. The relative speed of separation equals the relative speed of approach. Think of two billiard balls or ideal gas molecules. The "spring" gives back everything.

$e = 0$ — Perfectly Inelastic Collision

The objects stick together after impact. Their relative velocity of separation is zero. Think of two lumps of clay or a bullet embedding in a block. All the relative motion is lost.

$0 < e < 1$ — Partially Inelastic (Real World)

Every real collision falls here. A tennis ball against concrete might have $e \approx 0.7$ ; a basketball around $0.8$ . Some kinetic energy is lost as heat, sound, and permanent deformation.

Watch out

A common mistake: thinking $e$ depends on the masses of the colliding objects. It does not. $e$ is a property of the materials and the geometry of the surfaces in contact. A steel ball bearing has roughly the same $e$ whether it hits a small steel block or a large one.

Why "Along the Line of Impact" Matters

The definition includes the phrase "along the line of impact." In a glancing collision (like a pool ball hitting another at an angle), the velocities have components both along and perpendicular to the line joining the centers. The coefficient of restitution applies only to the velocity components along that line. The perpendicular components obey a different rule (usually, they are unchanged if friction is negligible).

A Quick Example …

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'PE converted to KE' during a bounce means $\frac{1}{2}mv_{up}^2=0.64\times\frac{1}{2}mv_{down}^2$ for the SAME bounce, i.e. $\left(\frac{v_{up}}{v_{down}}\right)^2=0.64$ , so the speed ratio (= coefficient of restitution e for a bounce off a fixed/immovable surface, $u_2=v_2=0$ ) is $e=\sqrt{0.64}=0.8$ .
Drop height 5 m gives the FIRST impact speed $v_0=\sqrt{2gh}=\sqrt{2\times10\times5}=10$ m/s. After each bounce the rebound speed is e times the impact speed, and (since the ball rises then falls back under gravity alone between bounces) the NEXT impact speed equals that same rebound speed. So impact speed before the n-th bounce, and rebound speed after it, both scale as $10\times e^{n-1}$ for the impact and $10\times e^n$ for the rebound. After the THIRD bounce, rebound speed $=10\times e^3=10\times0.512=5.12$ m/s. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. For an Inelastic collision, what is the value of coefficient of restitution (e)?$
(a) e = 0
(b) e = 1
(c) 0 < e < 1
(d) 0 > e > 1
›Reveal solutionSolution

For an inelastic collision, the coefficient of restitution e satisfies 0 < e < 1.

The coefficient of restitution is defined as e = (relative velocity of separation)/(relative velocity of approach) = (v2 − v1)/(u1 − u2). It measures how much kinetic energy is conserved in a collision. If e = 1, the collision is perfectly elastic (no kinetic energy lost, relative speed of approach equals relative speed of separation). If e = 0, the collision is perfectly inelastic (the bodies stick together after collision, no separ …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. For an elastic collision the coefficient of restitution$
(e) will be
(a) e > 1
(b) e < 1
(c) e = 1
(d) e = 0
›Reveal solutionSolution

The coefficient of restitution e is the ratio of relative speed of separation to relative speed of approach; for a perfectly elastic collision (KE conserved), e = 1.

e = (relative velocity of separation) / (relative velocity of approach)

For a perfectly elastic collision, kinetic energy is fully conserved and the relative speed of separation equals the relative speed of approach, giving e = 1.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. On dropping from a height of 64 cm, a spherical ball jumps up to 36 cm. The coefficient of restitution of collision is (A) 2/3 (B) 3/4 (C) 64/36 (D) 9/16$
›Reveal solutionSolution

$e=\sqrt{h_2/h_1}=\sqrt{36/64}=3/4$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. For perfectly elastic collision:$
(a) e = 1
(b) e > 1
(c) e < 1
(d) e = 0
›Reveal solutionSolution
For a perfectly elastic collision, e = 1.

The coefficient of restitution e = (relative velocity of separation)/(relative velocity of approach).

Perfectly elastic collision: kinetic energy conserved, e = 1. …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. In the case of perfectly elastic collision —$
(a) e = 1
(b) e > 1
(c) e < 1
(d) e = 0
›Reveal solutionSolution

For a perfectly elastic collision, e = 1.

The coefficient of restitution is $e = \dfrac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$ .

…

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Q.A ball of mass 100 g dropped on the ground from a height of 5 m bounces repeatedly. During every bounce, 64% of the potential energy (just before that bounce) is converted into kinetic energy (just after that bounce). Calculate the following:

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