Test Index

BITSAT 2016 Paper

Section: Physics

Question : 31 of 150

Marks: +1, -0

\theta_1

\theta_2

\theta_1+\theta_2

, is :

Solution:

Given

m_1=m_2

We will apply the principle of conservation of momentum in the mutually perpendicular

\text{dir}^n

Along x-axis

m_1 u_1 = m_1 v_1 \cos\theta + m_2 v_2 \cos\phi

u_1 = v_1 \cos\theta + v_2 \cos\phi

....(1)

Along y-axis

0 = m_1 v_1 \sin\theta - m_2 v_2 \sin\phi

0 = v_1 \sin\theta - v_2 \cos\phi

......(2)

Again for elastic collision, kinetic energy is conserved

\Rightarrow \;\; \frac{1}{2} m u_1^2 = \;\; \frac{1}{2} m v_1^2 + \;\; \frac{1}{2} m v_2^2

u_1^2 = v_1^2 + v_2^2

....(iii)

Squaring and adding (i) & (ii), we get

u_1^2 = v_1^2 (\cos^2\theta + \sin^2\theta) + v_2^2 (\cos^2\phi + \sin^2\phi) + 2 v_1 v_2 \cos\theta \cos\phi - 2 v_1 v_2 \sin\theta \sin\phi

u_1^2 = v_1^2 + v_2^2 + 2 v_1 v_2 \cos(\theta + \phi)

......(iv)

Using (iii) & (iv), we get

\cos(\theta + \phi) = 0 = \cos \;\; \frac{\pi}{2} \Rightarrow \theta + \phi = \;\; \frac{\pi}{2}

Note : This is a standard case of oblique collision.

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