Verified Solution • 2025–26 Session moderate

QJEEM24Ph3781: JEE Main 2024 Physics - Class 11 JEEM24Ph3781 Solution: Four particles of mass , have same momentum, respectively. The…

Four particles A,B,C,DA, B, C, D of mass m2,m,2m,4m\frac{m}{2}, m, 2 m, 4 m , have same momentum, respectively. The particle with maximum kinetic energy is :

Multiple Choice Options

  • A B
  • B C
  • C D
  • D A

Step-by-Step Verified Solution

The momentum pp of a particle is given by the product of its mass mm and its velocity vv , that is, p=m⋅vp = m \cdot v . For a given momentum, the relationship between mass and velocity can be understood as inversely proportional. This means that as the mass increases, the velocity decreases to maintain the same momentum, and vice versa.

The kinetic energy ( K.E.K.E. ) of a particle is given by the formula K.E.=12mv2K.E. = \frac{1}{2} m v^2 . This equation shows that the kinetic energy depends on both the mass of the particle and the square of its velocity.

Given that four particles A,B,C,DA, B, C, D have masses m2,m,2m,4m\frac{m}{2}, m, 2 m, 4 m , respectively, and all have the same momentum, we can assume the momentum of each particle to be pp . This common value of momentum allows us to express the velocity of each particle in terms of its mass and the common momentum pp . The velocity vv of each particle will be v=pmv = \frac{p}{m} .

Thus, for each particle, we can determine the velocity as follows:

  • For AA : vA=pm2=2pmv_A = \frac{p}{\frac{m}{2}} = \frac{2p}{m}

  • For BB : vB=pmv_B = \frac{p}{m}

  • For CC : vC=p2m=p2mv_C = \frac{p}{2m} = \frac{p}{2m}

  • For DD : vD=p4mv_D = \frac{p}{4m}

Now, substituting these velocities into the kinetic energy formula yields the kinetic energies for each particle:

  • K.E.A=12⋅m2⋅(2pm)2=12⋅m2⋅4p2m2=2p2mK.E._A = \frac{1}{2} \cdot \frac{m}{2} \cdot \left(\frac{2p}{m}\right)^2 = \frac{1}{2} \cdot \frac{m}{2} \cdot \frac{4p^2}{m^2} = \frac{2p^2}{m}

  • K.E.B=12⋅m⋅(pm)2=12⋅m⋅p2m2=p22mK.E._B = \frac{1}{2} \cdot m \cdot \left(\frac{p}{m}\right)^2 = \frac{1}{2} \cdot m \cdot \frac{p^2}{m^2} = \frac{p^2}{2m}

  • K.E.C=12⋅2m⋅(p2m)2=12⋅2m⋅p24m2=p24mK.E._C = \frac{1}{2} \cdot 2m \cdot \left(\frac{p}{2m}\right)^2 = \frac{1}{2} \cdot 2m \cdot \frac{p^2}{4m^2} = \frac{p^2}{4m}

  • K.E.D=12⋅4m⋅(p4m)2=12⋅4m⋅p216m2=p28mK.E._D = \frac{1}{2} \cdot 4m \cdot \left(\frac{p}{4m}\right)^2 = \frac{1}{2} \cdot 4m \cdot \frac{p^2}{16m^2} = \frac{p^2}{8m}

Comparing these kinetic energies, we see that the particle AA has the maximum kinetic energy, as it is inversely related to mass in this scenario, and AA has the least mass but the highest velocity squared component, thus maximizing its kinetic energy. Therefore, the correct answer is:

Option D: A

Related Learning Resources & Exam Preparation