Question

In Figure, two particles are launched from the origin of the coordinate system at time$\mathit{t}\mathbf{=}\mathbf{0}$. Particle 1 of mass${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{g}$is shot directly along the xaxis on a frictionless floor, with constant speed10.0m/s . Particle 2 of mass${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}$is shot with a velocity of magnitude20.0m/s, at an upward angle such that it always stays directly above particle 1. (a) What is the maximum height H_max reached by the com of the two-particle system? In unit-vector notation, (b) what are the velocity and (c) what are the acceleration of the com when the com reaches H_max

Short answer

(a) The maximum height $H_{max}$reached by the center of mass of the two-particle system is localid="1654241536744" $H_{max}=5.74\mathrm{m}$

(b) The unit vector notation of the velocity of the center of mass at $H_{max}$is $\underset{v_{cm}}{\to }=(10.0m/s)\hat{i}$

(c) The unit vector notation of the acceleration of the center of mass at$H_{max}$is$\underset{{\mathrm{a}}_{\mathrm{com}}}{\to }=(-3.68m/s^2)\hat{j}$

Step-by-step solution

Listing the given quantities

The mass of the particle 1 is $m_1=5.00\mathrm{g}=5.00\times {10}^{-3}\mathrm{kg}$

The initial velocity of the particle 1 is$v_1=10.0\mathrm{m}/\mathrm{s}$

The mass of the particle 2 is $m_2=3.00g=3.00\times {10}^{-3}\mathrm{kg}$

The velocity of the particle 2 is$\overrightarrow{v}=20.2\mathrm{m}/\mathrm{s}$

Understanding the concept of vector addition and center of mass

We can use the concept of vector addition theorem and kinematical equation and find the maximum height of particle 2. By using the expression of the center of mass we can find the center of mass of the maximum height of the system. To find velocity and acceleration of the center of mass of the system we can use the concept of motion of a system’s center of mass.

Formula:

$v^2=v_0^2‐2g{y}_{max}$ (i)

$\overrightarrow{R_{cm}}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}}{m_1+m_2}$(ii)

$M\overrightarrow{V_{cm}}=m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+...+m_n\overrightarrow{v_n}$(iii)

$M\overrightarrow{a_{cm}}=m_1\overrightarrow{a_1}+m_2\overrightarrow{a_2}+...+m_n\overrightarrow{a}$(iv)

(a) calculating the maximum height Hmax reached by the center of mass of the two-particle system

The maximum height reached by the center of mass of the two-particle system:

Particle 1 is moving along the x-axis. Hence $v_{1x}=10.0m/s$

Particle 1 is in x-y plane just above the first particle, hence the horizontal x component of particle 2 is,

$$\begin{gathered} V_{2x}=V_{1x} \\ =10.0m/s \end{gathered}$$

The vertical component of particle 2 is$V_{2y}$According to the Pythagoras theorem,

$$\begin{gathered} V_{2y}=\sqrt{V_2^2-V_{2x}^2} \\ =\sqrt{{(20.0m/s)}^2-{(10.0m/s)}^2} \\ =\sqrt{300} \\ 17.32m/s \end{gathered}$$

We can find the maximum height of particle 2 by using equation (i) as,

$$\begin{gathered} 0=V_{2y}^2-2g{y}_{max} \\ {Y}_{max}=\frac{V_{2y}^2}{2g} \\ =\frac{{(17.32\mathrm{m}/\mathrm{s})}^2}{2\times 9.8\mathrm{m}/{\mathrm{s}}^2} \\ =15.3\mathrm{m} \end{gathered}$$

From equation (ii), the expression for the center of mass of the system is,

$\overrightarrow{R_{cm}}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}}{m_1+m_2}$

We can apply this expression for the two-particle system, but particle 1 is moving along x direction only. Substitute the values of the given terms.

$$\begin{gathered} \overrightarrow{H_{max}}=\frac{m_1\left(0\right)+m_2\overrightarrow{y_{max}}}{m_1+m_2} \\ =\frac{3.00\times {10}^{-3}\mathrm{kg}\times 15.3\mathrm{m}}{5.00\times {10}^{-3}\mathrm{kg}+3.00\times {10}^{-3}\mathrm{kg}} \\ =5.74\mathrm{m} \end{gathered}$$

The maximum height $H_{max}$reached by the center of mass of the two-particle system is $H_{max}=5.74\mathrm{m}$

(b) unit vector notation of the velocity of the center of mass at Hmax

Particle 1 and Particle 2 have the same horizontal velocity components. At maximum height of the center of mass of the system, the vertical velocity component of particle 2 is zero. Thus, the motion of the system’s center of mass as given by equation (iii) is,

$$\begin{gathered} M\overrightarrow{v_{cm}}=m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+...+m_n\overrightarrow{v_n} \\ \left(m_1+m_2\right)\overrightarrow{v_{cm}}=m_1\overrightarrow{v_{1x}}+m_2\overrightarrow{v_{2x}} \\ \overrightarrow{v_{cm}}=\frac{m_1\overrightarrow{v_{1x}}+m_2\overrightarrow{v_{2x}}}{\left(m_1+m_2\right)} \end{gathered}$$

Substitute the values to find the value of the velocity of center of mass.

$$\begin{gathered} \underset{V_{cm}}{\to }=\frac{m_1(10.0m/s)\hat{j}+m_2(10.0m/s)\hat{i}}{(m_1+m_{2)}} \\ =\frac{(m_1+m_2)(10.0m/s)\hat{i}}{(m_1+m_{2)}} \\ =(10.0m/s)\hat{i} \end{gathered}$$

The unit vector notation of the velocity of the center of mass at $H_{max}$is

$\underset{V_{cm}}{\to }=(10.0m/s)\hat{j}$

(c) unit vector notation of the acceleration of the center of mass at Hmax

The unit vector notation of the acceleration of the center of mass at$H_{max}$

Particle 1 is moving with constant speed along the x-axis, but Particle 2 has downward acceleration. Then, the expression of the motion of the system’s center of mass as given by equation (iv) is,

$$\begin{gathered} M\overrightarrow{a_{cm}}=m_1\overrightarrow{a_1}+m_2\overrightarrow{a_2}+...+m_n\overrightarrow{a} \\ \left(m_1+m_2\right)\overrightarrow{a_{cm}}=m_1\left(0\right)+m_{2}g \\ {a}_{cm}=\frac{m_{2}g}{\left(m_1+m_2\right)} \end{gathered}$$

Substitute the values to find acceleration of center of mass.

$$\begin{gathered} a_{cm}=\frac{3.00\times {10}^{-3}kg\times 9.8m/s^2}{\left[(5.00\times {10}^{-3kg})+(3.00\times {10}^{-3}kg)\right]} \\ =(3.68m/s^2) \end{gathered}$$

This gives us the magnitude of acceleration. Since it is in downward direction, we can calculate acceleration aslocalid="1656146794786" $\overrightarrow{a_{com}}=(-3.68m/s^2)\hat{j}.$