A ball drops to the ground from a height of 2.0 m and then rebounds to a height of 1.5 m . Determine the coefficient of restitution.

Correct answer: B. 3 2 The velocity of the ball just before hitting the ground is v_1 = 2gh_1 The velocity of the ball just after rebounding from the ground is v_2 = 2gh_2 The coefficient of restitution e is given by the ratio of velocity of separation to the velocity of approach: e = (v_2)/(v_1) = (h_2)/(h_1) Given h_1 = 2.0 m and h_2 = 1.5 m e = (1.5)/(2.0) = (3)/(4) = 3 2

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Centre of Mass, Linear momentum, Collision

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Centre of Mass, Linear momentum, Collision

A ball drops to the ground from a height of 2.0 m and then rebounds...

A ball drops to the ground from a height of

2.0 \text{ m}

and then rebounds to a height of

1.5 \text{ m}

. Determine the coefficient of restitution.

Options

A $\dfrac{3}{4}$
B $\dfrac{\sqrt{3}}{2}$
C $\dfrac{4}{3}$
D $\dfrac{2}{\sqrt{3}}$

Correct answer

B. $\dfrac{\sqrt{3}}{2}$

Step-by-step solution

The velocity of the ball just before hitting the ground is

v_1 = \sqrt{2gh_1}

The velocity of the ball just after rebounding from the ground is

v_2 = \sqrt{2gh_2}

The coefficient of restitution

e

is given by the ratio of velocity of separation to the velocity of approach:

e = \dfrac{v_2}{v_1} = \sqrt{\dfrac{h_2}{h_1}}

Given

h_1 = 2.0 \text{ m}

and

h_2 = 1.5 \text{ m}

e = \sqrt{\dfrac{1.5}{2.0}} = \sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{2}

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