JEE MAIN - Physics (2018 (Offline) - No. 20)
In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If
the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative
velocity between the two particles, after collision, is :
$${{{v_0}} \over {\sqrt 2 }}$$
$${{v_0}} \over 4$$
$$\sqrt 2 {v_0}$$
$${{v_0}} \over 2$$
Explanation
From conservation of linear momentum,
mv0 = mv1 + mv2
or v0 = v1 + v2 ........(1)
According to the question,
Kf = $${3 \over 2}$$Ki
$$ \Rightarrow $$ $${1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2$$
$$ \Rightarrow $$ $$v_1^2 + v_2^2 = {3 \over 2}v_0^2$$
Using eq (1) $${\left( {{v_1} + {v_2}} \right)^2} = v_0^2$$
$$ \Rightarrow $$ $$v_1^2 + v_2^2 + 2{v_1}{v_2}$$ = $$v_0^2$$
$$ \Rightarrow $$ $$2{v_1}{v_2}$$ = $$v_0^2 - {3 \over 2}v_0^2$$ = $$ - {1 \over 2}v_0^2$$
Now, $${\left( {{v_1} - {v_2}} \right)^2}$$ = $${\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}$$
= $$v_0^2 - \left( { - v_0^2} \right)$$ = $$2v_0^2$$
$$\therefore$$ $${{v_1} - {v_2}}$$ = $$\sqrt 2 {v_0}$$
mv0 = mv1 + mv2
or v0 = v1 + v2 ........(1)
According to the question,
Kf = $${3 \over 2}$$Ki
$$ \Rightarrow $$ $${1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2$$
$$ \Rightarrow $$ $$v_1^2 + v_2^2 = {3 \over 2}v_0^2$$
Using eq (1) $${\left( {{v_1} + {v_2}} \right)^2} = v_0^2$$
$$ \Rightarrow $$ $$v_1^2 + v_2^2 + 2{v_1}{v_2}$$ = $$v_0^2$$
$$ \Rightarrow $$ $$2{v_1}{v_2}$$ = $$v_0^2 - {3 \over 2}v_0^2$$ = $$ - {1 \over 2}v_0^2$$
Now, $${\left( {{v_1} - {v_2}} \right)^2}$$ = $${\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}$$
= $$v_0^2 - \left( { - v_0^2} \right)$$ = $$2v_0^2$$
$$\therefore$$ $${{v_1} - {v_2}}$$ = $$\sqrt 2 {v_0}$$
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