Question

A collision occurs between a 2.00 kgparticle travelling with velocity ${\mathbf{v}}_{\mathbf{1}}\mathbf{=}\left(-4.00\frac{m}{s}\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\left(-5.00\frac{m}{s}\right)\stackrel{\mathbf{⏜}}{\mathbf{j}}$and a 4.00 kgparticle travelling with velocity ${\mathbf{v}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\frac{\mathbf{m}}{\mathbf{s}}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\frac{\mathbf{m}}{\mathbf{s}}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{j}}$. The collision connects the two particles. What then is their velocity in (a) unit-vector notation and as a (b) Magnitude and (c) Angle?

Short answer

a) Velocity in unit vector notation is $2.67\stackrel{⏜}{i}-3\stackrel{⏜}{j}$.

b) Magnitude of velocity is $4.01\frac{m}{s}$.

c) Angle is $-48.4°$.

Step-by-step solution

Step 1: Given

i) Mass of the one particle is 2.0 kg

ii) Mass of the other particle is 4.0 kg

iii) Velocity of the one particle is,$\overrightarrow{v_1}=-4\stackrel{⏜}{i}-5\stackrel{⏜}{j}$

iv) Velocity of the other particle is,$\overrightarrow{v_2}=6\stackrel{⏜}{i}-2\stackrel{⏜}{j}$

Determining the concept

Using the formula for the velocity vector of the center of mass, find the velocity vector of the center of mass of particles P and Q. From this, find its magnitude and direction.

Formula is as follow:

${\overrightarrow{v}}_{com}=\frac{\left({\overrightarrow{v}}_1{m}_1+{\overrightarrow{v}}_2{m}_2\right)}{\left(m_1+m_2\right)}$

Here, m₁, m₂ are masses, v₁, v₂ are velocities of masses m₁, m₂ and v_cm is velocity of centre of mass.

(a) Determine the velocity in unit vector notation

Use following formula to calculate velocity unit vector notation,

$$\begin{gathered} {\overrightarrow{v}}_{com}=\frac{\left({\overrightarrow{v}}_1{m}_1+{\overrightarrow{v}}_2{m}_2\right)}{\left(m_1+m_2\right)} \\ {v}_{com}=\frac{\left(-4\stackrel{⏜}{i}-5\stackrel{⏜}{j}\right)2+\left(6\stackrel{⏜}{i}-2\stackrel{⏜}{j}\right)4}{4+2} \\ {v}_{com}=2.66666\stackrel{⏜}{i}-3\stackrel{⏜}{j}\approx 2.67\stackrel{⏜}{i}-3\stackrel{⏜}{j} \end{gathered}$$

Hence, velocity in unit vector notation is $2.67\stackrel{⏜}{i}-3\stackrel{⏜}{j}$.

(b) Determine the magnitude of velocity

Magnitude of velocity is as follows:

$$\begin{gathered} v_{com}=\sqrt{2.{66666}^2+{\left(-3\right)}^2} \\ {v}_{com}=4.01\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, magnitude of velocity is $4.01\frac{m}{s}$.

(c) Determine the angle

Angle is,

$$\begin{gathered} \tan \theta =\frac{v_y}{v_x} \\ \tan \theta =-\frac{3}{2.66666} \\ \theta =-48.4° \end{gathered}$$

Hence,angle is$-48.4°$.

From the masses and velocities of objects, the magnitude and direction of the velocity of their center of mass can be found.