1. **Problem statement:** We analyze a bullet of mass $m=20\,g=0.02\,kg$ fired with velocity $V_0$ embedding into a wooden block of mass $M=1\,kg$ suspended by a string of length $l=1\,m$. After impact, the block and bullet move together with velocity $V_1$ and swing to an angle $\alpha=37^\circ$. We want to determine $V_0$ and analyze conservation laws. 2. **Physical principles and formulas:** - Conservation of linear momentum during collision: $$ (m+M)V_1 = mV_0 $$ - Mechanical energy conservation after collision (pendulum motion): $$ \text{Initial kinetic energy} = \text{Potential energy at max height} $$ - Height gained by pendulum center of mass: $$ h = l(1 - \cos\alpha) $$ - Gravitational potential energy reference at lowest point $G_1$. 3. **Step 1: Conservation during collision** - Momentum conserved, kinetic energy not necessarily conserved. - Linear momentum before impact: $$ p_i = mV_0 $$ - Linear momentum after impact: $$ p_f = (M+m)V_1 $$ - Equate: $$ (M+m)V_1 = mV_0 \implies V_1 = \frac{m}{M+m} V_0 $$ 4. **Step 2: Mechanical energy just after impact** - Kinetic energy just after impact: $$ E_{k1} = \frac{1}{2}(M+m)V_1^2 = \frac{1}{2}(M+m)\left(\frac{m}{M+m}V_0\right)^2 = \frac{1}{2}\frac{m^2}{M+m}V_0^2 $$ 5. **Step 3: Mechanical energy at highest point $G_2$** - Potential energy at height $h$: $$ E_{p2} = (M+m)gh = (M+m)g l (1 - \cos\alpha) $$ - At highest point, kinetic energy is zero, so total mechanical energy is potential energy. 6. **Step 4: Equate energies to find $V_0$** - Mechanical energy conserved after collision: $$ E_{k1} = E_{p2} $$ - Substitute: $$ \frac{1}{2}\frac{m^2}{M+m}V_0^2 = (M+m)g l (1 - \cos\alpha) $$ - Solve for $V_0$: $$ V_0 = \sqrt{\frac{2(M+m)^2 g l (1 - \cos\alpha)}{m^2}} = \frac{M+m}{m} \sqrt{2 g l (1 - \cos\alpha)} $$ 7. **Step 5: Verify conservation laws** - Linear momentum is conserved during collision. - Kinetic energy is not conserved during collision because some energy converts to heat, deformation, etc. **Final answers:** - $V_1 = \frac{m}{M+m} V_0$ - $V_0 = \frac{M+m}{m} \sqrt{2 g l (1 - \cos\alpha)}$ - Linear momentum conserved, kinetic energy not conserved during collision.