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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

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 ms1100\text{ ms}^{-1}

B

80 ms180\text{ ms}^{-1}

C

40 ms140\text{ ms}^{-1}

D

120 ms1120\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 ms1p = 20\text{ kg ms}^{-1}
  5. Calculate Velocity: vs=pms=200.2=100 ms1v_s = \frac{p}{m_s} = \frac{20}{0.2} = 100\text{ ms}^{-1}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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