A shell of mass $200 ext{ g}$ is ejected from a gun of mass $4 ext{ kg}$ by an explosion that generates $1.0

Question

A shell of mass $200	ext{ g}$ is ejected from a gun of mass $4	ext{ kg}$ by an explosion that generates $1.05	ext{ kJ}$ of energy. The initial velocity of the shell is:

Accepted Answer

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 $m_s, v_s$ be the mass and velocity of the shell, and $m_g, v_g$ for the gun.
   $$ P_i = 0 \implies P_f = m_s v_s + m_g v_g = 0 $$
   The magnitudes of momentum are equal: $p = m_s v_s = m_g v_g$ [Class 11 Physics, Ch 4, Sec 4.7, Eq 4.12].
3. **Energy Equation:** Total Energy $E = K_s + K_g$.
   Using the relation $K = \frac{p^2}{2m}$:
   $$ E = \frac{p^2}{2m_s} + \frac{p^2}{2m_g} = \frac{p^2}{2} \left( \frac{1}{m_s} + \frac{1}{m_g} ight) $$
4. **Substitute Values:**
   $m_s = 0.2	ext{ kg}$, $m_g = 4	ext{ kg}$, $E = 1050	ext{ J}$.
   $$ 1050 = \frac{p^2}{2} \left( \frac{1}{0.2} + \frac{1}{4} ight) = \frac{p^2}{2} (5 + 0.25) $$
   $$ 1050 = \frac{p^2}{2} (5.25) \implies 2100 = 5.25 p^2 \implies p^2 = 400 $$
   $$ p = 20	ext{ kg ms}^{-1} $$
5. **Calculate Velocity:**
   $$ v_s = \frac{p}{m_s} = \frac{20}{0.2} = 100	ext{ ms}^{-1} $$

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