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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
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B
π4+α
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C
π4α
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D
π2
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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

$$\displaystyle
\tan\theta=\dfrac{m
v
_{0}\cos\alpha}{m
u
_{0}\cos\alpha}=1
$$
$$
$$
$$
\Rightarrow  \theta\ =\pi/4
$$

102424_31868_ans.PNG

Physics

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