Given :
Distance of the falling rock before striking the inclined plane, h = [tex]32[/tex] m
Slope, [tex]$\alpha = 40^\circ$[/tex]
The coefficient of restitution = [tex]0.2[/tex]
To find : The velocity of the rock after the impact.
The velocity of the given rock after the impact is [tex]16.70[/tex] m/sec
The potential drop of the rock is = Change in its kinetic energy
[tex]i.e.[/tex] [tex]mgh= \frac{1}{2}mv^2[/tex]
[tex]\Rightarrow v_A=\sqrt{2gh}[/tex]
[tex]\Rightarrow v_A=\sqrt{2\times 10 \times 32}[/tex]
[tex]\Rightarrow v_A=25.29[/tex] m/sec
The velocity of the rock just before collision in terms of co-ordinate axes is as :
[tex]$v_A=-v_A \sin 40^\circ\ \widehat{i} - v_A \cos 40^\circ\ \widehat{j} $[/tex]
[tex]=(-16.25\ \widehat{i}-19.37 \ \widehat{j})[/tex] m/sec
The collision of the ball with the plane is an oblique impact, with the line of action perpendicular to the plane. The co-efficient of restitution applies perpendicular to the incline and the momentum of the ball is conserved along the incline.
Therefore, the co-efficient of restitution applies along the y-axis :
[tex]$e=\frac{[(v_{B_y})_2-(v_{A_y})_2]}{[(v_{A_y})_1-(v_{B_y})_1]}$[/tex]
[tex]$(v_{B_y})_1=(v_{B_y})_2=0$[/tex] (Since. the incline is still)
[tex]$\Rightarrow 0.2=\frac{-(v_{A_y})_2}{-19.37}$[/tex]
[tex]$\Rightarrow (v_{A_y})_2=(0.2)(19.37)$[/tex]
[tex]$\Rightarrow (v_{A_y})_2=3.874$[/tex] m/sec
Now conserving momentum along the [tex]x[/tex] direction, we get
[tex]$m_A(v_{A_x})_1=m_A(v_{A_x})_2$[/tex]
[tex]$(v_{A_x})_1=(v_{A_x})_2 = -16.25 \ m/sec$[/tex]
[tex]\therefore[/tex] Velocity of the rock after the collision is :
[tex]$v=\sqrt{(v_{A_x})_2^2+(v_{A_y})_2^2}$[/tex]
[tex]$v=\sqrt{(-16.25)^2+(3.874)^2}$[/tex]
[tex]v=16.70[/tex] m/sec
So, after the collision, the velocity of the rock is 16.70 m/sec.
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