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NEET PHYSICSEasy

A particle of mass 4M4M at rest splits into two particles of mass MM and 3M3M. The ratio of the kinetic energies of mass MM and 3M3M would be:

A

3:1

B

1:4

C

1:1

D

1:3

Step-by-Step Solution

  1. Conservation of Momentum: Since the initial particle is at rest, the initial total momentum is zero. In the absence of external forces, the total momentum of the system is conserved during the split (explosion/decay) [Class 11 Physics, Ch 6, Sec 6.4, Eq 6.18a]. Pinitial=Pfinal=0\vec{P}_{initial} = \vec{P}_{final} = 0 p1+p2=0    p1=p2\vec{p}_1 + \vec{p}_2 = 0 \implies \vec{p}_1 = -\vec{p}_2 The magnitudes of momentum for both particles are equal: p1=p2=pp_1 = p_2 = p.
  2. Relate Kinetic Energy to Momentum: Kinetic energy (KK) is related to momentum (pp) and mass (mm) by the formula: K=p22mK = \frac{p^2}{2m}
  3. Calculate Ratio: For the first particle (mass MM): K1=p22MK_1 = \frac{p^2}{2M} For the second particle (mass 3M3M): K2=p22(3M)K_2 = \frac{p^2}{2(3M)} Ratio K1K2=p22Mp26M=p22M×6Mp2=3\text{Ratio } \frac{K_1}{K_2} = \frac{\frac{p^2}{2M}}{\frac{p^2}{6M}} = \frac{p^2}{2M} \times \frac{6M}{p^2} = 3 Thus, the ratio is 3:13:1.
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