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NEET PHYSICSMedium

A shell of mass 200 g200\text{ g} is ejected from a gun of mass 4 kg4\text{ kg} by an explosion that generates 1.05 kJ1.05\text{ kJ} of energy. The initial velocity of the shell is:

A

100 ms−1100\text{ ms}^{-1}

B

80 ms−180\text{ ms}^{-1}

C

40 ms−140\text{ ms}^{-1}

D

120 ms−1120\text{ ms}^{-1}

Step-by-Step Solution

  1. Identify Principles: The explosion is an internal force, so the total linear momentum of the system (gun + shell) is conserved. The energy generated converts into the kinetic energy of both the shell and the gun.
  2. Conservation of Momentum: Let ms,vsm_s, v_s be the mass and velocity of the shell, and mg,vgm_g, v_g for the gun. Pi=0  ⟹  Pf=msvs+mgvg=0P_i = 0 \implies P_f = m_s v_s + m_g v_g = 0 The magnitudes of momentum are equal: p=msvs=mgvgp = m_s v_s = m_g v_g [Class 11 Physics, Ch 4, Sec 4.7, Eq 4.12].
  3. Energy Equation: Total Energy E=Ks+KgE = K_s + K_g. Using the relation K=p22mK = \frac{p^2}{2m}: E=p22ms+p22mg=p22(1ms+1mg)E = \frac{p^2}{2m_s} + \frac{p^2}{2m_g} = \frac{p^2}{2} \left( \frac{1}{m_s} + \frac{1}{m_g} \right)
  4. Substitute Values: ms=0.2 kgm_s = 0.2\text{ kg}, mg=4 kgm_g = 4\text{ kg}, E=1050 JE = 1050\text{ J}. 1050=p22(10.2+14)=p22(5+0.25)1050 = \frac{p^2}{2} \left( \frac{1}{0.2} + \frac{1}{4} \right) = \frac{p^2}{2} (5 + 0.25) 1050=p22(5.25)  ⟹  2100=5.25p2  ⟹  p2=4001050 = \frac{p^2}{2} (5.25) \implies 2100 = 5.25 p^2 \implies p^2 = 400 p=20 kg ms−1p = 20\text{ kg ms}^{-1}
  5. Calculate Velocity: vs=pms=200.2=100 ms−1v_s = \frac{p}{m_s} = \frac{20}{0.2} = 100\text{ ms}^{-1}
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