JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 13)
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
1
Explanation
By conservation of linear momentum :
(0.1)(3$$\widehat i$$) + (0.1)(5$$\widehat j$$) = (0.1)(4)($$\widehat i$$ + $$\widehat j$$) + (0.1)$$\overrightarrow v $$
$$ \Rightarrow $$ $$\overrightarrow v = - \widehat i + \widehat j$$
$$ \therefore $$ $$\left| {\overrightarrow v } \right|$$ = $$\sqrt 2 $$
KEB = $${1 \over 2} \times 0.1 \times {\left( {\sqrt 2 } \right)^2}$$ = $${1 \over {10}}$$ J
$$ \therefore $$ x = 1
(0.1)(3$$\widehat i$$) + (0.1)(5$$\widehat j$$) = (0.1)(4)($$\widehat i$$ + $$\widehat j$$) + (0.1)$$\overrightarrow v $$
$$ \Rightarrow $$ $$\overrightarrow v = - \widehat i + \widehat j$$
$$ \therefore $$ $$\left| {\overrightarrow v } \right|$$ = $$\sqrt 2 $$
KEB = $${1 \over 2} \times 0.1 \times {\left( {\sqrt 2 } \right)^2}$$ = $${1 \over {10}}$$ J
$$ \therefore $$ x = 1
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