JEE MAIN - Physics (2017 (Offline) - No. 26)
A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The
collision is head on, and elastic. The ratio of the de-Broglie wavelengths $${\lambda _A}$$ to $${\lambda _B}$$ after the collision is:
$${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 3}$$
$${{{\lambda _A}} \over {{\lambda _B}}} = 2$$
$${{{\lambda _A}} \over {{\lambda _B}}} = {2 \over 3}$$
$${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 2}$$
Explanation
From question, mA = M; mB = $${m \over 2}$$
uA = V and uB = 0
Let after collision velocity of A = V1 and
velocity of B = V2
Applying law of conservation of momentum,
mu = mv1 + $$\left( {{m \over 2}} \right){v_2}$$
$$ \Rightarrow $$ 24= 2v1 + v2 ........(i)
By law of collision
$$e = {{{v_2} - {v_1}} \over {u - 0}}$$
$$ \Rightarrow $$ u = v2 - v1 ..........(ii)
[As collision is elastic, e = 1]
using eqns (i) and (ii)
v1 = $${4 \over 3}$$ and v2 = $${4 \over 3}v$$
We know, de-Broglie wavelength $$\lambda $$ = $${h \over p}$$
$$ \therefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{{P_B}} \over {{P_A}}}$$ = $${{{m \over 2} \times {4 \over 3}u} \over {m \times {4 \over 3}}}$$ = 2
uA = V and uB = 0
Let after collision velocity of A = V1 and
velocity of B = V2
Applying law of conservation of momentum,
mu = mv1 + $$\left( {{m \over 2}} \right){v_2}$$
$$ \Rightarrow $$ 24= 2v1 + v2 ........(i)
By law of collision
$$e = {{{v_2} - {v_1}} \over {u - 0}}$$
$$ \Rightarrow $$ u = v2 - v1 ..........(ii)
[As collision is elastic, e = 1]
using eqns (i) and (ii)
v1 = $${4 \over 3}$$ and v2 = $${4 \over 3}v$$
We know, de-Broglie wavelength $$\lambda $$ = $${h \over p}$$
$$ \therefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{{P_B}} \over {{P_A}}}$$ = $${{{m \over 2} \times {4 \over 3}u} \over {m \times {4 \over 3}}}$$ = 2
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