Question

Agroup of particles of total mass $\mathbf{35}\mathbf{}\mathit{k}\mathit{g}$has a total kinetic energy of $\mathbf{340}\mathbf{}\mathit{J}$. The kinetic energy relative to the center of mass is $\mathbf{85}\mathbf{}\mathit{J}$. What is the speed of the center of mass?

Short answer

The speed of the center of mass is$\mathbf{4}\mathbf{.}\mathbf{93}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$

Step-by-step solution

Identification of the given data

• The total mass is $m=35kg$
• Total kinetic energy is $K_T=340J$
• Kinetic energy relative to the center of mass is $K=85J$

Concept of the center of mass of the system

The center of mass of the system is calculated by considering the placement of the of all the objects acting in the system.

Determination of the speed of the center of mass

According to the conservation of energy principle,

${\mathit{K}}_{\mathbf{T}}\mathbf{+}\mathit{K}\mathbf{=}\mathit{E}$

Where,

$K_T$= Total kinetic energy

K = Kinetic energy relative to the center of mass

Potential Energy, localid="1653897038194" role="math" $\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{{\mathbf{v}}_{\mathbf{c}}}^{\mathbf{2}}$

Substitute Equation (2) in (1),

$$\begin{gathered} K_T+K=\frac{1}{2}m{v_c}^2 \\ {v_c}^2=2·\left(\frac{K_T+K}{m}\right) \\ =2·\left(\frac{340+85}{35}\right) \\ {v_c}^2=24.29 \\ {v}_c=4.93m/s \end{gathered}$$

Hence,the speed of the center of mass is localid="1653897138661" $\mathbf{4}\mathbf{.}\mathbf{93}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$