Question

23. A ball of mass m hits the floor with a speed v making
an angle of incidence 0 with the normal. The coefficient
of restitution is e. Find the speed of the reflected ball and
the angle of reflection of the ball.
0
e'
Figure 9-W16​

Answer 1

AnswEr :

• Speed of reflected ball is V √(e² cos² θ + sin θ)

• Angle made with normal is $\small\sf{ \tan}^{ - 1}$ tanθ/ e .

GivEn :

• Velocity of approach(V) is the normal component of velocity(Vn) .

• Coefficient of restitution is e.

• Angle of incidence is θ with respect to the normal .

To find :

• Speed of reflected ball and angle of reflection of the ball = ?

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★ Concept to be used :

For solving this question you need to know the concept of coefficient of restitution and velocity of approach(seperation) !!

Coefficient of restitution is the ratio of the final to initial velocity between any 2 objects or bodies after they collide .

It is given by,

$\dag \large \ { \boxed{ \rm{v2-v1= - e(u 2 - u 1)}}}$

Where, coefficient of restitution satisfies 0≤e≤1.

Meanwhile, velocity of approach is given as

$\dag \ { \boxed{\rm \: v = v_{left} - v_{right}}}$

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Solution :

As, velocity of approach is the normal component of velocity here,

➝ Vn = V cos θ

➝ Vn =eVn [As per velocity of approach ]

➝ Vn = e V cos θ ⠀⠀⠀⠀⠀⠀−[eqn 1]

Hence, the tangential velocity(Vt) wouldn't change !!

So,

➝ Vt = V sin θ ⠀⠀⠀⠀⠀⠀−[eqn 2]

Now, Speed of reflected ball will be,

➝ √Vn² + Vt²

➝ √(e V cos θ)² + (V sin θ)²

➝ V√(e² cos² θ + sin² θ) [V is taken as common factor ]

Further,

Angle with normal would be,

➝ θ = $\small\sf{ \tan}^{ - 1}$ Vt / Vn

➝ θ = $\small\sf{ \tan}^{ - 1}$ Vsin θ/eVcos θ[From eqn 1 and 2]

➝ θ = $\small\sf{ \tan}^{ - 1}$ tanθ/ e

Therefore,

• Speed of reflected ball is V √(e² cos² θ + sin θ)

• Angle made with normal is $\small\sf{ \tan}^{ - 1}$ tanθ/ e .

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