Question

Two blocks of masses \[10\,kg\] and $4\,kg$ are connected by a spring of negligible mass and are placed on a frictionless horizontal surface. An impulse gives a speed of $14m/s$ to the heavier block in the direction of the lighter block. then, the velocity of the Centre of mass is:
A) $30\,m{s^{ - 1}}$
B) $20\,m{s^{ - 1}}$
C) $10\,m{s^{ - 1}}$
D) $5\,m{s^{ - 1}}$

Answer 1

Hint: The mass velocity equation center is the sum of the momentum of each particle mass times velocity divided by the total mass of the system, and we can also calculate the missing quantity in any numerical if any of these two quantities are given by using this formula.

Formula used:
Velocity of center of mass
${V_{CM}} = \dfrac{{{m_1}{v_1} + {m_1}{v_1}}}{{{m_1} + {m_2}}}$
${v_1}$ is the velocity of the first particle,
\[\;{v_2}\;\] is the velocity of the second particle,
$m$ is the total mass of the system.

Complete step by step solution:
Given by,
Mass of the block ${m_1} = 10\,kg$
Mass of the block ${m_2} = 4\,kg$
Velocity of mass ${m_1}$, ${v_1} = 14\,m/s$
Velocity of mass \[{m_2}\] ,${v_2} = 0$
If we assume that all of the system's mass is located in the mass center of the system.
According to the velocity of center of mass,
$\Rightarrow$ ${V_{CM}} = \dfrac{{{m_1}{v_1} + {m_1}{v_1}}}{{{m_1} + {m_2}}}$
Substituting the given value in above equation,
$\Rightarrow$ ${V_{CM}} = \dfrac{{10 \times 14 + 0}}{{10 + 4}}$
On simplifying,
$\Rightarrow$ ${V_{CM}} = \dfrac{{140}}{{14}}$
We get, ${V_{CM}} = 10\,m/s$

Hence, The option C is the correct answer.

Note: When Velocity is an indicator of how much time an object takes to reach a directional destination. If a rigid body is considered, the center of mass may or may not be identical to the geometric center. It is regarded as a reference point for many other mechanical calculations.