A particle of mass is projected from the ground with an initial speed at an angle 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

Question

A particle of mass $m$ is projected from the ground with an initial speed $u_{0}$ at an angle $α$ 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 $u_{0}$ . The angle that the composite system makes with the horizontal immediately after the collision is [2013]

Smart Achievers · Accepted Answer

(1)

$Height, h = \frac{u_{0}^{2} \sin^{2} α}{2 g}$

$using v^{2} - u^{2} = 2 g h$

$v_{1}^{2} - u_{0}^{2} = 2 (- g) [\frac{u_{0}^{2} \sin^{2} α}{2 g}]$

$\Rightarrow v_{1}^{2} = u_{0}^{2} (1 - \sin^{2} α) = u_{0}^{2} \cos^{2} α$

$\Rightarrow v_{1} = u_{0} \cos α$

Applying conservation of linear momentum in Y-direction

$2 m v \sin θ = m v_{1} = m u_{0} \cos α ...(i)$

Applying conservation of linear momentum in X-direction

$2 m v \cos θ = m u_{0} \cos α ...(ii)$

$Dividing (i) and (ii) we get$

$\tan θ = 1 \Rightarrow θ = 45 ° = \frac{π}{4}$

Solution

1 $\frac{π}{4}$

2 $\frac{π}{4} + α$

3 $\frac{π}{2} - α$

4 $\frac{π}{2}$

Want Us to Email you About Special Offers & Updates?

Solution

1 π4

2 π4+α

3 π2-α

4 π2

Want Us to Email you About Special Offers & Updates?

1 $\frac{π}{4}$

2 $\frac{π}{4} + α$

3 $\frac{π}{2} - α$

4 $\frac{π}{2}$