Question

Two particles A and B of mass of 10 g and 20 g respectively fall vertically. At a given time, the speed of particle A is 12 m/s and that of B is 15 m/s. Find the total linear momentum of the system of the two particles.

Answer 1

Given

1. Mass of particle A ($m_a$) = $10g=10\times {10}^{-3}kg.$
2. Mass of the particle B ($m_b$) = $20g=20\times {10}^{-3}kg$
3. Velocity of particle A ($v_a$) = $12m/s$
4. Velocity of particle B ($v_b$) = $15m/s$

Calculation of the total linear momentum of the system of the two particles

1. Momentum of a body is defined as the product of mass and velocity.
2. Momentum of a body is given by:$p=mv$.
3. If there are two or more bodies then the total linear momentum of the system will be the sum of the momentum of all the bodies.
4. Total linear momentum = Momentum of body A + Momentum of body B.
5. Total linear momentum = $m_a{v}_a+m_b{v}_b.$
6. Total linear momentum =$10\times {10}^{-3}\times 12+20\times {10}^{-3}\times 15$.
7. Total linear momentum = ${10}^{-3}(120+300)=420\times {10}^{-3}=0.42kgm{s}^{-1}$

Hence, the total linear momentum of the system of the two particles is$\mathbf{0}\mathbf{.}\mathbf{42}\mathbf{}\mathit{k}\mathit{g}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.