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

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWork, Energy, and Power5 Marks
Question
A block of mass $2.0kg$ moving at $2.0m/s$ collides head on with another block of equal mass kept at rest.
  1. Find the maximum possible loss in kinetic energy due to the collision.
  2. If the actual loss in kinetic energy is half of this maximum, find the coefficient of restitution.
✓

Answer

Mass of block $=2 \mathrm{~kg}$ and speed $=2 \mathrm{~m} / \mathrm{s}$ Mass of $2^{\text {nd }}$ block $=2 \mathrm{~kg}$. Let final velocity of $2^{\text {nd }}$ block $=\mathrm{v}$
a. Using law of conservation of momentum.
$2 \times 2=(2+2) \mathrm{v}$
$\Rightarrow \mathrm{v}=1 \mathrm{~m} / \mathrm{s}$
$\therefore$ Loss in K.E. in inelastic collision
$=\Big(\frac{1}{2}\Big)\times2\times(2)^2\text{v}-\Big(\frac{1}{2}\Big)(2+2)\times(1)^2$
$=4-2=2\text{J}$
b.$\text{Actual loss}=\frac{\text{Maximum loss}}{2}=1\text{J}$ $=\Big(\frac{1}{2}\Big)2\times(2)^2-
\Big(\frac{1}2{}\Big)2\times\text{v}_1^2+\Big(\frac{1}2{}\Big)\times2\times\text{v}_2^2=1$
$\Rightarrow4-\big(\text{v}_1^2+\text{v}^2_2\big)=1$
$\Rightarrow4-\frac{(1+\text{e}^2)\times4}{2}=1$
$\Rightarrow2(1+\text{e}^2)=3$
$\Rightarrow1+\text{e}^2=\frac{3}{2}$
$\Rightarrow\text{e}^2=\frac{1}{2}$
$\Rightarrow\text{e}=\frac{1}{\sqrt2}$

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