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Work, Energy, and Power — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsWork, Energy, and Power5 Marks
Question
Two identical balls A and B undergo a perfectly elastic two dimensional collision. Initially A is moving with a speed of 10ms-1 and B is at rest. Due to collision A is scattered through angle of 30°. What are the speed of A and B after the collision?
✓

Answer

The balls A and B are identical
$\therefore$  masses are same, $\theta=30^\circ$
$\therefore\phi=90^\circ-30^\circ=60^\circ$
Initially
$\text{v}_1=10\text{ms}^{-1}$
and $\text{v}_2=0$
$\therefore$ According to Law of conservation of momentum for x- component'.
$\text{u}=\text{v}_1\cos30^\circ+\text{v}_2\cos60^\circ$
$10=\text{v}_1\times\frac{\sqrt{3}}{2}+\text{v}_2\times\frac{1}{2}$
$\therefore\sqrt{3}\text{v}_1+\text{v}_2=20$
For y-component,
$0=\text{v}_1\sin30^\circ-\text{v}_2\sin60^\circ$
$0=\text{v}_1\times\frac{1}{2}-\text{v}_2\times\frac{\sqrt{3}}{2}$
$\text{v}_1=\sqrt{3}\text{v}_2$
Form eq. (i) and (ii) we get
$\sqrt{3}\times\sqrt{3}\text{v}+\text{v}_2=20$
$\Rightarrow3\text{v}_2+\text{v}_2=20$
$\Rightarrow4\text{v}_2=20$
$\therefore\text{v}_2=5\text{ ms}^{-1}$
From eq. (iii)
$\text{v}_1=\sqrt{3}\times5=5\sqrt{3}\text{ ms}^{-1}$
Here, the velocities of A and B are $5\sqrt{3}\text{ ms}^{-1}$ and 5ms-1 respectively.

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