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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2018 · 8 April · Physics · Center of Mass, Momentum and Collision

It is found that if a neutron suffers an elastic collinear...

It is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss of its energy is Pd{P}_{d}, while for its similar collision with a carbon nucleus at rest, the fractional loss of energy is Pc{P}_{c}. The values of Pd{P}_{d} and Pc{P}_{c} are respectivelyIt is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss ofIt is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss ofIt is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss ofIt is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss of

Options

  1. A0,10, 1
  2. B0.89,0.280 . 89, 0 . 28
  3. C0.28,0.890 . 28, 0 . 89
  4. D0,00, 0

Correct answer

B. 0.89,0.280 . 89, 0 . 28

Step-by-step solution

It is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss of
Since, the linear momentum is conserved,& u=V1+2V2u = {V}_{1} + 2{V}_{2}.

e=1=V2−V1ue = 1 = \dfrac{{V}_{2} - {V}_{1}}{u}~(where& e⇒e ⇒the~coefficient of restitution),& u=V2−V1u = {V}_{2} - {V}_{1}

V2=2u3;V1=−u3{V}_{2} = \dfrac{2u}{3}; {V}_{1} = - \dfrac{u}{3}. Initial energy,~=12u2= \dfrac{1}{2}u^{2}. Final energy,~=12×1×u29=u218= \dfrac{1}{2} \times 1 \times \dfrac{u2}{9} = \dfrac{u2}{18}. Fractional change =u22−u218u22=0.88= \dfrac{\dfrac{u^{2}}{2} - \dfrac{u^{2}}{18}}{\dfrac{u^{2}}{2}}= 0.88

It is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss of

u=V1+12V2⇒1=V2−V1u⇒u=V2−V1u = {V}_{1} + 12{V}_{2}⇒ 1 = \dfrac{{V}_{2} - {V}_{1}}{u}⇒ u = {V}_{2} - {V}_{1}
⇒V2=2u13⇒V1=−11u13⇒ {V}_{2} = \dfrac{2u}{13}⇒ {V}_{1} = - \dfrac{11u}{13}.
Change in energy,=u2−(11u13)2×100=0.294= u^{2} - {\left(\dfrac{11u}{13}\right)}^{2} \times 100= 0.294
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