Define angular momentum; exploding 10 kg projectile

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.