Question

# A particle of mass $$m$$ is projected from the ground with an initial speed $$u_{0}$$at an angle $$\alpha$$with the horizontal. $$A$$t 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

A
π4
B
π4+α
C
π4α
D
π2

Solution

## The correct option is A $$\dfrac{\pi}{4}$$Velocity of the particle performing projectile motion at highest point$$=V_{1}=V_{0}cos\alpha$$Velocity particle thrown vertically upwards at the position of collision$$\displaystyle=v_{2}^{2}=u_{0}^{2}-2g\dfrac{u^{2}\sin^{2}\alpha}{2g}=v_{0}\cos\alpha$$So, from conservation of momentum$$\displaystyle\tan\theta=\dfrac{mv_{0}\cos\alpha}{mu_{0}\cos\alpha}=1$$$$\Rightarrow \theta\ =\pi/4$$Physics

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