Question

A particle of mass m is projected from the ground with an initial speed u, at an angle $\{\alpha}$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another
identical particle, which was thrown vertically upward from the ground with the same initial speed u.
The angle that the composite system makes with the horizontal immediately after the collision is :
1)$\frac {\pi}{4}$
2)$\frac {\pi}{4}$ + a
3)$\frac {\pi}{4}$ - a
4)none.

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Answer 1

Formulas used :-

› v² - u² = 2gh

› $\sf H_{max}$ = $\dfrac{u²sin²θ}{2g}$

› tan θ = $\dfrac{y}{x}$

Solution :-

• Velocity of the particle thrown vertically upwards, $\sf v_2²$ = $\sf u_o²$ - 2g ($\dfrac{u_o²sin²α}{2g}$)

• $\sf v_2²$ = $\sf u_o²$ - $\sf u_o²$sin²α

• $\sf v_2$ = $\sf u_o²$$\sqrt{1-sin²α}$

• $\sf v_2$ = $\sf u_o$ cos α

• Velocity of the particle in a projectile motion, $\sf v_1$ = $\sf u_o$ cos α

As, after collision they meet at a specific point (see attachment), let's make the components.

Now, let's divide the velocities in order to find the slope.

• tan θ = 1 = 45°

• tan θ = $\dfrac{π}{4}$

$\bold{Hope\;it \; helps\;!}$

Answer 2

Answer:

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