--- title: "Define angular momentum; exploding 10 kg projectile" subject: "Applied Mechanics (Dynamics)" chapter: 3 key: "semester_3/applied_mechanics_dynamics:angular-momentum-define-explosion-10kg" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/angular-momentum-define-explosion-10kg site: BCT Engineering --- # Define angular momentum; exploding 10 kg projectile Applied Mechanics (Dynamics) · Chapter 3: System of particles - Asked in: 2078 Bhadra, 2+4 marks - Video: [Momentum Explosions](https://www.youtube.com/watch?v=MkJommMrIJc) — Physics with Professor Matt Anderson - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/angular-momentum-define-explosion-10kg - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3 ## Question Define angular momentum for a system of particles. A 10 kg projectile is moving with a velocity of 30 m/s when it explodes into two fragments A and B weighing 2.5 kg and 7.5 kg respectively, knowing that immediately after the explosion, fragments A and B travel in the directions defined respectively by $\theta_A = 45^\circ$ and $\theta_B = 30^\circ$, determine the velocity of each fragment. [Figure: 10 kg projectile moving right at $V_0 = 30$ m/s; after explosion fragment A (labelled 1.5 kg in the figure) moves at $\theta_A$ above the horizontal and fragment B (7.5 kg) at $\theta_B$ below the horizontal.] ## Short answer Angular momentum about a fixed point O is H_O = Σ r_i × m_i v_i; its derivative equals the external moment. Momentum conservation: 10(30) = 2.5 v_A cos45° + 7.5 v_B cos30°, 0 = 2.5 v_A sin45° – 7.5 v_B sin30°. Solving: v_A = 62.1 m/s at 45° above, v_B = 29.3 m/s at 30° below. ## Answer #### Angular momentum of a system of particles The angular momentum (moment of momentum) of a system of $n$ particles about a fixed point O is the vector sum of the moments of the momenta of the particles: $$ \vec H_O=\sum_{i=1}^{n}\vec r_i\times m_i\vec v_i $$ About the mass centre G, using velocities relative to G or absolute ones, $\vec H_G=\sum\vec r'_i\times m_i\vec v'_i$. The two are related by $\vec H_O=\vec{\bar r}\times m\vec{\bar v}+\vec H_G$. Its rate of change equals the moment of the external forces: $\sum\vec M_O=\dot{\vec H}_O$. Unit: kg m$^2$/s. #### Exploding projectile **Data:** $m=10$ kg, $v_0=30$ m/s, $m_A=2.5$ kg, $m_B=7.5$ kg (the masses given in the question; the "1.5 kg" in the figure label is treated as a misprint because $2.5+7.5=10$). $\theta_A=45^\circ$ above, $\theta_B=30^\circ$ below the horizontal. Linear momentum is conserved. **x-direction:** $$ 10(30)=2.5v_A\cos45^\circ+7.5v_B\cos30^\circ $$ $$ 300=1.7678v_A+6.4952v_B\qquad(1) $$ **y-direction:** $$ 0=2.5v_A\sin45^\circ-7.5v_B\sin30^\circ\ \Rightarrow\ v_A=2.1213v_B\qquad(2) $$ Using (2) in (1): $$ 300=3.75v_B+6.4952v_B=10.245v_B\ \Rightarrow\ v_B=29.28\ \text{m/s},\quad v_A=62.12\ \text{m/s} $$ **Answer: $v_A=62.1$ m/s at $45^\circ$ above the horizontal; $v_B=29.3$ m/s at $30^\circ$ below the horizontal.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the angular momentum of a system about O defined? A. Sum of m_i v_i B. Sum of r_i times m_i v_i C. Total mass times mean speed D. Sum of m_i r_i Answer: B. H_O = sum of r_i x m_i v_i, the moments of the particle momenta about O. (Key point: Definition) 2. How are H_O and H_G related? A. H_O = H_G B. H_O = r_bar x m v_bar - H_G C. H_O = r_bar x m v_bar + H_G D. H_O = m v_bar + H_G Answer: C. H_O = r_bar x m v_bar + H_G: the mass-centre term plus the angular momentum about G. (Key point: Relation H_O and H_G) 3. What equals the rate of change of H_O? A. Sum of the external forces B. The total linear momentum C. The kinetic energy D. The moment of external forces about O Answer: D. Sum M_O = dH_O/dt. (Key point: Rate of change) 4. What are the fragment speeds when the masses are 2.5 kg and 7.5 kg? A. v_A = 62.1 m/s, v_B = 29.3 m/s B. v_A = 29.3 m/s, v_B = 62.1 m/s C. v_A = 30 m/s, v_B = 30 m/s D. v_A = 124 m/s, v_B = 58.6 m/s Answer: A. Momentum conservation gives v_A = 2.1213 v_B and 300 = 10.245 v_B, so v_B = 29.28 m/s, v_A = 62.12 m/s. 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