During the head on collision of two bodies of masses 1 kg and 2 kg- the maximum energy of deformation is 1003J- If before collision- the masses are moving in the same direction- then their velocity of approach in m-s before the collision is

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Standard X

Physics

Mechanical Advantage

During the he...

Question

During the head on collision of two bodies of masses $1 k g$ and $2 k g$ , the maximum energy of deformation is $\frac{100}{3} J$ . If before collision, the masses are moving in the same direction, then their velocity of approach (in $m / s$ ) before the collision is

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Solution

When deformation is maximum, both the particles are moving with same velocity. So, applying momentum conservation,

$m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{1}^{'} + m_{2} v_{1^{'}}$

$v_{1}^{'} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

Applying energy conservation,
$\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

$= \frac{1}{2} (m_{1} + m_{2}) (v_{1^{'}})^{2} + Δ U$ deformation

$\Rightarrow Δ U$ deformation $= \frac{1}{2} \frac{m_{1} m_{2}}{(m_{1} + m_{2})} \times (v_{1} - v_{2})^{2} = \frac{100}{3}$

$\Rightarrow v_{1} - v_{2} = 10 m / s$
Final answer: 10

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