JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 12)
Particle A of mass m1
moving with velocity $$\left( {\sqrt3\widehat i + \widehat j} \right)m{s^{ - 1}}$$ collides with another particle B of mass m2
which is at rest initially. Let $$\overrightarrow {{V_1}} $$
and $$\overrightarrow {{V_2}} $$
be the velocities of particles A and B after collision
respectively. If m1
= 2m2
and after
collision $$\overrightarrow {{V_1}} = $$$$\left( {\widehat i + \sqrt 3 \widehat j} \right)$$ , the angle between $$\overrightarrow {{V_1}} $$ and $$\overrightarrow {{V_2}} $$ is :
collision $$\overrightarrow {{V_1}} = $$$$\left( {\widehat i + \sqrt 3 \widehat j} \right)$$ , the angle between $$\overrightarrow {{V_1}} $$ and $$\overrightarrow {{V_2}} $$ is :
105o
15o
-45o
60o
Explanation
Given m1
= 2m2
So let, m2 = m and m1 = 2m
From momentum conservation
$${\overrightarrow p _i}$$ = $${\overrightarrow p _f}$$
$$ \Rightarrow $$ (2m)$$\left( {\sqrt 3 \widehat i + \widehat j} \right)$$ + 0 = 2m$$\left( {\widehat i + \sqrt 3 \widehat j} \right)$$ + m$${\overrightarrow V _2}$$
$$ \Rightarrow $$ $${\overrightarrow V _2}$$ = 2$$\left( {\sqrt 3 \widehat i + \widehat j} \right)$$ - 2$$\left( {\widehat i + \sqrt 3 \widehat j} \right)$$
$$ \Rightarrow $$ $${\overrightarrow V _2}$$ = $$\left( {2\sqrt 3 - 2} \right)\widehat i - \widehat j\left( {2\sqrt 3 - 2} \right)$$
= $$2\left( {\sqrt 3 - 1} \right)\left( {\widehat i - \widehat j} \right)$$
Also given after collision $$\overrightarrow {{V_1}} = \left( {\widehat i + \sqrt3\widehat j} \right)m{s^{ - 1}}$$
For angle between $${\overrightarrow V _1}$$ & $${\overrightarrow V _2}$$,
cos $$\theta $$ = $${{{{\overrightarrow V }_1}.{{\overrightarrow V }_2}} \over {\left| {{{\overrightarrow V }_1}} \right|\left| {{{\overrightarrow V }_2}} \right|}}$$
= $${{2\left( {\sqrt 3 - 1} \right) \times 1 - 2\left( {\sqrt 3 - 1} \right) \times \sqrt 3 } \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}$$
= $${{2\left( {\sqrt 3 - 1} \right)\left( {1 - \sqrt 3 } \right)} \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}$$
= $${{\left( {1 - \sqrt 3 } \right)} \over {2\sqrt 2 }}$$
$$ \Rightarrow $$ $$\theta $$ = 105o
So let, m2 = m and m1 = 2m
From momentum conservation
$${\overrightarrow p _i}$$ = $${\overrightarrow p _f}$$
$$ \Rightarrow $$ (2m)$$\left( {\sqrt 3 \widehat i + \widehat j} \right)$$ + 0 = 2m$$\left( {\widehat i + \sqrt 3 \widehat j} \right)$$ + m$${\overrightarrow V _2}$$
$$ \Rightarrow $$ $${\overrightarrow V _2}$$ = 2$$\left( {\sqrt 3 \widehat i + \widehat j} \right)$$ - 2$$\left( {\widehat i + \sqrt 3 \widehat j} \right)$$
$$ \Rightarrow $$ $${\overrightarrow V _2}$$ = $$\left( {2\sqrt 3 - 2} \right)\widehat i - \widehat j\left( {2\sqrt 3 - 2} \right)$$
= $$2\left( {\sqrt 3 - 1} \right)\left( {\widehat i - \widehat j} \right)$$
Also given after collision $$\overrightarrow {{V_1}} = \left( {\widehat i + \sqrt3\widehat j} \right)m{s^{ - 1}}$$
For angle between $${\overrightarrow V _1}$$ & $${\overrightarrow V _2}$$,
cos $$\theta $$ = $${{{{\overrightarrow V }_1}.{{\overrightarrow V }_2}} \over {\left| {{{\overrightarrow V }_1}} \right|\left| {{{\overrightarrow V }_2}} \right|}}$$
= $${{2\left( {\sqrt 3 - 1} \right) \times 1 - 2\left( {\sqrt 3 - 1} \right) \times \sqrt 3 } \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}$$
= $${{2\left( {\sqrt 3 - 1} \right)\left( {1 - \sqrt 3 } \right)} \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}$$
= $${{\left( {1 - \sqrt 3 } \right)} \over {2\sqrt 2 }}$$
$$ \Rightarrow $$ $$\theta $$ = 105o
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