A body of mass $4m$ is lying in the $xy$-plane at rest. It suddenly explodes into three pieces. Two pieces eac

Question

A body of mass $4m$ is lying in the $xy$-plane at rest. It suddenly explodes into three pieces. Two pieces each of mass $m$ move perpendicular to each other with equal speeds $v$. The total kinetic energy generated due to the explosion is:

Accepted Answer

1. **Conservation of Momentum:** The body is initially at rest, so the total initial momentum is zero. According to the Law of Conservation of Momentum, the vector sum of the momenta of the fragments after the explosion must be zero.
   $$ \vec{P}_{initial} = 0 \implies \vec{P}_{final} = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 $$
   (Refer to NCERT Class 11, Chapter 5, Section 5.7 ).
2. **Analyze Fragment Momenta:**
   - Piece 1: Mass $m$, speed $v$. Let $\vec{p}_1 = mv \hat{i}$.
   - Piece 2: Mass $m$, speed $v$ (perpendicular). Let $\vec{p}_2 = mv \hat{j}$.
   - Piece 3: Mass $m_3 = 4m - m - m = 2m$. Its momentum $\vec{p}_3$ must balance the sum of the first two.
   $$ \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) $$
   Magnitude: $|p_3| = \sqrt{(mv)^2 + (mv)^2} = \sqrt{2}mv$.
3. **Calculate Speed of Third Piece ($v_3$):**
   $$ p_3 = m_3 v_3 \implies \sqrt{2}mv = (2m)v_3 $$
   $$ v_3 = \frac{\sqrt{2}v}{2} = \frac{v}{\sqrt{2}} $$
4. **Calculate Total Kinetic Energy:**
   $$ K_{total} = K_1 + K_2 + K_3 $$
   $$ K_1 = \frac{1}{2}mv^2, \quad K_2 = \frac{1}{2}mv^2 $$
   $$ K_3 = \frac{1}{2}(2m)(v_3)^2 = m\left(\frac{v}{\sqrt{2}}\right)^2 = m\left(\frac{v^2}{2}\right) = \frac{1}{2}mv^2 $$
   $$ K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = \frac{3}{2}mv^2 $$
   (Refer to NCERT Class 11, Chapter 6, Section 5.4 for Kinetic Energy ).

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