Question

The particle 'P' shown in figure makes a collision with floor and then performs projectile motion of range 15m. The coefficient of restitution for the collision is $e = \frac{\sqrt{x}}{y}$, where x and y are coprime. The value of x+y is ______. (g = 10 m/s²)

Answer 1

7

Answer 2

1. Pre-collision velocity:
   Since the particle is launched from the ground with speed $u = 20\,\text{m/s}$ at $60^\circ$, when it lands back on the ground (ignoring air resistance), its vertical speed just before impact is

   $$|v_y| = u\sin60 = 20\cdot\frac{\sqrt3}{2} = 10\sqrt3\,\text{m/s}.$$

   Its horizontal component remains

   $$v_x = u\cos60 = 20\cdot\frac{1}{2} = 10\,\text{m/s}.$$

2. Effect of collision:
   On collision with the floor, the horizontal component remains unchanged. The vertical component after collision becomes

   $$v_{y,\text{after}} = e\,(10\sqrt3)\; \text{(upward)}.$$

3. Range after collision:
   The time of flight after collision is given by

   $$T = \frac{2\,v_{y,\text{after}}}{g} = \frac{2e(10\sqrt3)}{10} = 2e\sqrt3.$$

   The horizontal range $R$ is

   $$R = v_x\,T = 10\,(2e\sqrt3) = 20e\sqrt3.$$

   Given that $R = 15\,\text{m}$, we have

   $$20e\sqrt3 = 15 \quad\Rightarrow\quad e = \frac{15}{20\sqrt3} = \frac{3}{4\sqrt3}.$$

   Rationalize the denominator:

   $$e = \frac{3\sqrt3}{4\cdot3} = \frac{\sqrt3}{4}.$$

4. Determining $x+y$:
   Since $e = \frac{\sqrt{x}}{y}$ and we obtained $e = \frac{\sqrt3}{4}$, we identify $x = 3$ and $y = 4$. Thus,

   $$x+y = 3+4 = 7.$$