Question

Two blocks of masses 10 kg and 30 kg are placed along a vertical line. If the first block is raised through a height of 7 cm by what distance should the second mass be moved to raise the center of mass by 1 cm. A)1cm up B) 2cm down C) 1cn down D) 2cm up

Answer 1

Let the second block move by a distance x (in cm) upward to make the center of mass move by 1 cm upward. Upward motion is considered as positive here.

The displacement of the center of mass of a system of two particles of masses $\sf{m_1}$ and $\sf{m_2}$ having individual displacements $\sf{x_1}$ and $\sf{x_2}$ respectively is,

$\longrightarrow\sf{\bar x=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}}$

Here,

• $\sf{m_1=10\ kg}$

• $\sf{m_2=30\ kg}$

• $\sf{x_1=7\ cm}$

• $\sf{x_2=x\ cm}$

• $\sf{\bar x=1\ cm}$

Then,

$\longrightarrow\sf{1=\dfrac{10\times7+30x}{10+30}}$

$\longrightarrow\sf{1=\dfrac{70+30x}{40}}$

$\longrightarrow\sf{70+30x=40}$

$\longrightarrow\sf{30x=-30}$

$\longrightarrow\sf{\underline{\underline{x=-1\ cm}}}$

It means the second block has to move by 1 cm downward. That's why the negative sign.