Question

A small particle elastically strikes a fixed smooth
sphere of 4 cm radius, with a speed of 5 m/s. The
impact parameter for the collision is 2√3cm.
Calculate the angle of deflection of the particle.
(1) 60° (2) 30° (3) 45° (4) 75°​

Answer 1

Answer:

Angle of deviation of the ball with respect to its initial direction is

$\theta = 60^o$

Explanation:

As we know that the impact parameter is 2sqrt3

so we can find the angle made by incident velocity with radius vector

so we have

$sin\theta = \frac{d}{R}$

$sin\theta = \frac{2\sqrt3}{4}$

$\theta = sin^{-1}\frac{\sqrt3}{2}$

so we have

$\theta = 60^o$

now angle of deviation is given as

$\delta = \pi - 2i$

$\delta = 180 - 2(60)$

$\delta = 60^o$

#Learn

Topic : Elastic Collision

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