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

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWork, Energy, and Power5 Marks
Question
A ball of mass m, moving with a speed $2v_0$, collides inelastically $(e > 0)$ with an identical ball at rest. Show that:
For a general collision, the angle between the two velocities of scattered balls is less than $90°$
✓

Answer

Consider the diagram shows general collision: (image not complet) Let angle between the $p_1$ and $p_2$ is $\theta$ By the law of conservation of momentum.$\vec{\text{p}}=\vec{\text{p}_1}+\vec{\text{p}_2}$
In inelastic collision, some part of KE lost in the from of heat, deshaping etc.$\therefore\ \text{KE}_\text{i}>\text{KE}_1+\text{KE}_2$
$\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{m(0)}^2>\frac{1}{2}\text{mv}^2_1+\frac{1}{2}\text{mv}^2_2$
$\frac{\text{p}^2}{2\text{m}}>\frac{\text{p}^2_1}{2\text{m}}+\frac{\text{p}^2_2}{2\text{m}}$
$\therefore\ \vec{\text{p}}^2>\vec{\text{p}}^2_1+\vec{\text{p}}^2_2\ ....(\text{iii})$
If $\text{p}^2=\text{p}^2_1+\text{p}^2_2$ then angle between $p_1$ and $p_2$ is $90^\circ$ So, equation (iii) is true when angle between $p_1, p_2$ than $90^\circ$ or acute as shown in also.

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