Double collision in projectile

Note: I have used some terms in the solution which i am defining

x = initial angle of projection

u = initial velocity of projection

e = coefficient of restitution

R= Horizontal Range of projectile thrown initially

Lets Consider That We Have Thrown The Ball With An Initial Velocity u at an

angle x with the horizontal.

Now The Important point to be noted is that if it strikes the VERTICAL WALL

PERPENDICULARLY then it must be moving horizontally at that particular

moment . Hence The collision MUST TAKE PLACE AT HIGHEST POINT .

As We know

at the topmost point of trajectory the particle only has horizontal component of

velocity as it does not change throughout the flight .

Lets Suppose the coefficient of restitution at the pair of surfaces is e

After the collision with wall it will rebound with a velocity eucosx

Now this will become the case of horizontal projection from a height .

we can find out the velocity with which the projectile will strike the ground first

time .

We can Apply Third equation of kinematics to see that(In vertical direction)

v = usinx

Also we can find out the horizontal distance covered by particle using second

equation of kinematics

x=eR/2 1.....

Velocity in horizontal direction will remain same as eucosx .

Now when the ball will collide the with the floor the line of impact will be

vertical . so the ball will rebound with a velocity eusinx in vertical direction and

velocity eucosx will remain unchanged as it will be perpendicular to the line of

impact .

It will be a case of normal projectile on ground.

It is easy to see that that projectile will now start at an angle x with the

horizontal .

Applying formula of range of projectile

y = R*e^2 2.........

In Order to reach the point of projection

The total Horizontal distance to be covered by the particle = R/2

Hence

1......+ 2....... = R/2

Solving the about equation

We get a quadratic in e

e = -1 or 0.5

Neglecting Negative value

e= 0.5 Nice Problem!

Total ascend distance,

$\displaystyle s = \frac{v_x \cdot v_y}{g}$

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After collision 1,

$\displaystyle v_{1-x}= e\cdot v_x,\; v_{1-y} = v_y$

$\displaystyle t_1 = \frac{v_y}{g}$

$\Rightarrow s_1 = v_{1-x} \cdot t_1 = \dfrac {e\cdot v_x \cdot v_y}{g}$

---

After collision 2,

$\displaystyle v_{2-x}= e\cdot v_x,\; v_{2-y} = e\cdot v_y$

$\displaystyle t_2 = \frac{e\cdot v_y}{g}$

$\displaystyle\Rightarrow s_2 = v_{2-x} \cdot t_2 = \dfrac {e\cdot v_x \cdot e\cdot v_y}{g}$

---

Now,

$\displaystyle s = s_1+s_2$

$\displaystyle\Rightarrow \dfrac{v_x \cdot v_y}{g} = \dfrac {e\cdot v_x \cdot v_y}{g} + \dfrac {e\cdot v_x \cdot e\cdot v_y}{g}$

Simplifying, $\displaystyle 2e^2 + e - 1 = 0$

Or, $\displaystyle e = \frac{-1\pm \sqrt{1+8}}{4} = \frac {2}{4}$

$∴\displaystyle\boxed{e = 0.5}$