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JEE Advanced 2015 · Paper 1, Physics question 20

A particle of unit mass is moving along the xx-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (aa and U0U_0 are constants). Match the potential energies in column I to the corresponding statement(s) in column II.

Column I

  • A
    U1(x)=U02[1−(xa)2]2U_1(x)=\dfrac{U_0}2\left[1-\left(\dfrac xa\right)^2\right]^2
  • B
    U2(x)=U02(xa)2U_2(x)=\dfrac{U_0}2\left(\dfrac xa\right)^2
  • C
    U3(x)=U02(xa)2exp⁡[−(xa)2]U_3(x)=\dfrac{U_0}2\left(\dfrac xa\right)^2\exp\left[-\left(\dfrac xa\right)^2\right]
  • D
    U4(x)=U02[xa−13(xa)3]U_4(x)=\dfrac{U_0}2\left[\dfrac xa-\dfrac13\left(\dfrac xa\right)^3\right]

Column II

  • P
    The force acting on the particle is zero at x=ax=a.
  • Q
    The force acting on the particle is zero at x=0x=0.
  • R
    The force acting on the particle is zero at x=−ax=-a.
  • S
    The particle experiences an attractive force towards x=0x=0 in the region ∣x∣<a|x|<a.
  • T
    The particle with total energy U04\dfrac{U_0}4 can oscillate about the point x=−ax=-a.
Mark every match for each entry
PQRST
A
B
C
D
Answer and worked solution

Official answer: A → P, Q, R, T; B → Q, S; C → P, Q, R, S; D → P, R, T

Use F=−dUdxF=-\dfrac{dU}{dx}, writing u=x/au=x/a.

U1=U02(1−u2)2U_1=\frac{U_0}2(1-u^2)^2. F=2U0au(1−u2)F=\dfrac{2U_0}{a}u(1-u^2), which is zero at x=0,±ax=0,\pm a: P, Q, R. For 0<x<a0<x<a, F>0F>0 (away from 0), so it is not attractive. U1(−a)=0U_1(-a)=0 is a minimum, and U1(0)=U02>U04U_1(0)=\frac{U_0}2>\frac{U_0}4, so a particle with energy U04\frac{U_0}4 oscillates about x=−ax=-a: T.

U2=U02u2U_2=\frac{U_0}2u^2. F=−U0auF=-\dfrac{U_0}{a}u: zero only at 0 (Q) and always towards 0 (S).

U3=U02u2e−u2U_3=\frac{U_0}2u^2e^{-u^2}. U3′∝u(1−u2)e−u2U_3'\propto u(1-u^2)e^{-u^2}: zero at 0,±a0,\pm a (P, Q, R). For ∣x∣<a|x|<a the force points towards 0 (S). x=±ax=\pm a are maxima, so there is no oscillation about −a-a.

U4=U02(u−u33)U_4=\frac{U_0}2\left(u-\frac{u^3}3\right). U4′∝1−u2U_4'\propto1-u^2: zero at ±a\pm a (P, R), not at 0. x=−ax=-a is a minimum with U=−U03U=-\frac{U_0}3, and the barrier at x=ax=a is +U03>U04+\frac{U_0}3>\frac{U_0}4, so it oscillates about −a-a (T).

Answer: A → P, Q, R, T; B → Q, S; C → P, Q, R, S; D → P, R, T

Topics:Potential energy and conservative forcesSimple harmonic motion
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