A cannon of mass 10 × 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 2.1 × 103 kg shell horizontally with an initial velocity of 550 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity (in m/s)of a shell fired horizontally by this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)

Engineering

Physics

Conservation of Energy

Newtons Second Law of Motion

Momentum and Energy Conservation for System of Particles

Question

A cannon of mass 10 × 10³ kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 2.1 × 10³ kg shell horizontally with an initial velocity of 550 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity (in m/s)of a shell fired horizontally by this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)

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Energy released = $\frac{1}{2}$ (2.1) × 10³ × 550²
Also energy relased = $\frac{1}{2} \times (\frac{10 \times 10^{3} \times 2.1 \times 10^{3}}{10 \times 10^{3} + 2.1 \times 10^{3}})$ (u + v)²
⇒    (u + v) = 550 × 1.1
       10 × 10³v = 2.1 × 10³u   (By momentum conservation)
       u + 0.21u = 550 × 1.1
       u = = 500 m/s

$u = \frac{550 \times 1.1}{1.21} = 500 m / s$