A body A, of mass m = 0.1 kg has an initial velocity of 3$\widehat i$ ms-1 . It collides elastically with another body, B of the same mass which has an initial velocity of 5$\widehat j$ ms-1. After collision, A moves with a velocity $\overrightarrow v = 4\left( {\widehat i + \widehat j} \right)$. The energy of B after collision is written as ${x \over {10}}$. The value of x is ___________.
Answer (integer)
1
Solution
By conservation of linear momentum :
<br><br>(0.1)(3$\widehat i$) + (0.1)(5$\widehat j$) = (0.1)(4)($\widehat i$ + $\widehat j$) + (0.1)$\overrightarrow v$
<br><br>$\Rightarrow$ $\overrightarrow v = - \widehat i + \widehat j$
<br><br>$\therefore$ $\left| {\overrightarrow v } \right|$ = $\sqrt 2$
<br><br>KE<sub>B</sub> = ${1 \over 2} \times 0.1 \times {\left( {\sqrt 2 } \right)^2}$ = ${1 \over {10}}$ J
<br><br>$\therefore$ x = 1
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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