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Question

A particle of mass m is projected with a speed u from the ground at angle is θ=π3 w.r.t. horizontal x-axis. When it has reached its maximum height, it collides completely in elastically with another particle of the same mass and velocity ui^. The horizontal distance covered by the combined mass before reaching the ground is:


A

33u28g

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B

22u2g

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C

5u28g

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D

32u24g

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Solution

The correct option is A

33u28g


Step 1: Given data and assumptions.

Mass of the particle=m

Projectile angle, θ=π3

Initial velocity =u

Step 2: Finding the horizontal velocity of the combined mass.

We know that,

According to the law of conservation of momentum,

Pi=Pf

Where, Pi is initial momentum, and Pf is final momentum.

mu+mucosθ=m+mvmu+mucosθ=2mv [As it collides elastically, a mass will stick to the other mass]

Where, m is the mass of the particle, u is the initial velocity of the particle, and v is the final velocity.

v=u1+cos602 cos60°=12

v=34u ….i

Step 3: Find the time required to reach the maximum height

Since,

Vertical maximum range,

H=12gt2

Where, t is time to reach the maximum height and g is the acceleration due to gravity.

t=2Hg ……ii

Step 4: Finding the horizontal distance traveled by the combined mass

Since the horizontal range after collision=vt

=34u×2Hg [ Fromi,ii]

=34u×2u2sin2602g2 H=u2sin2θ2g

=34u234g

=33u28g

Hence, option A is Correct.


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