Question

Example 5.5 Two identical billiard balls
strike a rigid wall with the same speed but
at different angles, and get reflected without
any change in speed, as shown in Fig. 5.6.
What is (i) the direction of the force on the
wall due to each ball? (ii) the ratio of the
magnitudes of impulses imparted to the
balls by the wall?

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Answer 1

$\bigstar\huge\boxed{\mathtt{\red{Solution:}}}$

Fig A :

A billiard ball of mass m moving with speed v strikes the wall at an angle 0 degree

We need to find the direction of force on the wall due to each ball

In order to do that ,first we need to find the change in momentum

As we know that ,

$\implies\mathtt{F=\dfrac{\Delta P}{t}}$

We know that ,

$\implies\mathtt{\Delta P = P_f - P_i }$

• Initial momentum = mv
• Final momentum = - mv (as the ball moves in the negative x direction)

$\implies\mathtt{\Delta P = (-mv)-mv }$

$\implies\mathtt{\Delta P = -2mv }$

Now substituting the value of ΔP in the above formula,

$\implies\mathtt{F=\dfrac{-2mv}{t}}$

Now lets find the impulse imparted by the ball on the wall ,

we know that ,

$\implies\mathtt{I= F.t}$

But ,

$\implies\mathtt{\Delta P= F.t}$

Hence from the above two equations we can conclude that ,

$\implies\mathtt{\Delta P_a= I_a= -2mv}$

$\implies\mathtt{ I_a= -2mv}$

---------------------------------------

Fig B:

A billiard ball of mass m moving with speed v strikes the wall at an angle 30 degree from the horizontal

We need to find the direction of force on the wall due to each ball

In order to do that ,first we need to find the change in momentum

As we know that ,

$\implies\mathtt{F=\dfrac{\Delta P}{t}}$

We know that ,

$\implies\mathtt{\Delta P = P_f - P_i }$

• Initial momentum = mv cosθ

• Final momentum = - mv cosθ(as the ball moves in the negative x direction)

$\implies\mathtt{\Delta P = (-mv cos\theta)-mvcos\theta }$

$\implies\mathtt{\Delta P = -2mvcos30 }$

$\implies\mathtt{\Delta P = -2mv\dfrac{\sqrt{3} }{2} }$

$\implies\mathtt{\Delta P = -\sqrt{3} mv }$

Now substituting the value of ΔP in the above formula,

$\implies\mathtt{F=\dfrac{-\sqrt{3} mv}{t}}$

Now lets find the impulse imparted by the ball on the wall ,

$\implies\mathtt{\Delta P_b= I_b= -\sqrt{3} mv}$

$\implies\mathtt{ I_b= -\sqrt{3} mv}$

Ratio of magnitude of impulses

$\implies\mathtt{\dfrac{I_a}{I_b} =\dfrac{2mv}{\sqrt{3}mv } }$

$\implies\mathtt{\pink{\dfrac{I_a}{I_b} =\dfrac{2}{\sqrt{3} }}}$