Answer to Question #142559 in Physics for Mason Earl

Question

Stuck on hw. A 50kg firework is shot at 35m/s before exploding, it splits into 2 smaller 25kg pieces. How fast to the smaller pieces move, ignoring air resistance

Answer 1

The intial momentum of the firework is:

$$p_i = mv_i = 50kg\times 35m/s = 1750\space kg\cdot m/s$$

According to the momentum conservation law, 2 smaller pieces should have the same momentum in total (in the direction of initial movment):

$$p_f = m_1v_1 + m_2v_2 = p_i$$

where $m_1 = m_2\equiv m_s = 25kg$ is the mass of one piece and $v_1$, $v_2$ are their velocities. Thus:

$$v_1 + v_2 = \dfrac{p_i}{m_s} = \dfrac{1750\space kg\cdot m/s}{25kg} = 70m/s$$

If we assume the conservation of kinetic energy (no external forces), then, due to the symmetry will be:

$$v_1 = v_2 = \dfrac{70m/s}{2} = 35m/s$$

Answer. Both velocities are 35 m/s.