A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is…

A particle of unit mass is moving along the x-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 (a and U0 are constants). Match the potential energies in column I to the corresponding statement(s) in column II.
  Column I   Column II
A. U1x=U021-xa22 P. The force acting on the particle is zero at x = a
B. U2x=U02xa2 Q. The force acting on the particle is zero at x = 0 .
C. U3x=U02xa2exp⁡-xa2 R. The force acting on the particle is zero at x= -a
D. U4x=U02xa-13xa3 S. The particle experiences an attractive force towards x = 0 in the region x<a
    T. The particle with total energy U04 can oscillate about the point x= -a
  1. a-s,t;b-s;c-s,t;d-t;
  2. a-s;b-r,s;c-r,s,t;d-s,t;
  3. a-r,s;b-q;c-q,r;d-t;
  4. a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;

Solution

U1x=U021-x2a22
F=-U0221-x2a2-2xa2
=2U0a4a2-x2x
F=2U0a4xa-xa+x
F=0 , at x=0, a, -a
x=-a, U=0,x=0, U=U02⇒ Oscillate when total ME is less then U02
A→P,Q,R,T
B- U0x=U02xa2
F=-U0xa2
F=0, x=0
B→Q,S
C- U3x=U02xa2exp⁡-x2a2
F=-U022xa2exp⁡-x2a2+x2a2exp⁡-x2a2-2xa2
=-U022xa4exp⁡-x2a2a2-x2
=U0a4ex2a2 xx+ax-a
x=0, x=-a, x=+a, F=0
x=0, U=0 even function hence minima
C→P,Q,R,S
D- U4x=U02xa-x33a3
F=-U0a1a-x2a3=U02a3x2-a2 
x=+a, x=-a, F=0
x=-a, U=-U03, x=+a, U=+U03
Hence oscillate about x=-a if T.M.E<U03
D→P,R,T !

Asked in: JEE Advanced 2015 (Paper 1)

Practice more Work Power Energy questions on Aicharya