Question

Solve the previous problem if the coefficient of restitution is e.
Use θ=45°, e=34 and h=5 m.

Answer 1

Explanation:

Step 1:

It is given that the plane has the inclination angle, $\theta=45^{\circ}$, From the height (h) of 5 m, a ball falls on the inclined plane. Then, the restitution coefficient is $e=3 / 4$.

$v=\sqrt{2 g \times 5}$

The inclined plane is struck by the velocity of the ball and it is shown as  

$=10 \mathrm{m} / \mathrm{s}$

Step 2:

After the collision, an angle β is made by the ball with the horizontal.

The velocity of the horizontal component, $10 \cos 45^{\circ}$ remains the same.

Though, in perpendicular direction to the plane, the velocity after the collision will be: $v 1=e \times 10 \sin 45^{\circ}$  

$=(3.75) 2 \mathrm{m} / \mathrm{s}$

Step 3:

Likewise, $v 2=52 \mathrm{m} / \mathrm{s}$

Now, $u=v 12+v 22=28.125+50=8.83 \mathrm{m} / \mathrm{s}$

 From the wall, the angle of reflection is shown as,

$\beta=\tan -1(3.752) 52=37^{\circ}$

Projection has the angle $\alpha=90-(\vartheta+\beta)$

$\Rightarrow \alpha=90^{\circ}-\left(45^{\circ}+37^{\circ}\right)=8^{\circ}$

Step 4:

After the collision, the balls fall in a place whose distance be L.

$x=L \cos \vartheta$

Projection of angle $(\alpha)=-8^{\circ}$

Now, $Y=x \tan \alpha-g \times 2 \sec 2 \alpha 2 u 2$

$\Rightarrow-L \sin \theta=L \cos \theta \times \tan 8^{\circ}-g 2 L 2 \cos 2 \theta \sec 280^{\circ}(u) 2$

$\Rightarrow-\sin 45^{\circ}=\cos 45^{\circ} \times \tan 8^{\circ}-10 \cos 245 \sec 28^{\circ}(8.83) 2$

When the above equation is solved, we obtain:

$L=18.5 \mathrm{m}$