Question

A particle of mass m is kept on smooth cube of mass m and side l.Cube starts moving i constant velocity v,s displacement of center of mass along horizontal direction when particle hits the ground formula is

Answer 1

Answer:

Displacement of center of mass of the system when particle hits the ground is given as

$d = \frac{v}{2}\sqrt{\frac{2l}{g}}$

Explanation:

As we know that the particle will drop from its top to bottom

So here as per the formula of free fall

$l = \frac{1}{2}gt^2$

so the time interval is given as

$t = \sqrt{\frac{2l}{g}}$

Now velocity of center of mass in horizontal direction is given as

$v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$

$v_[cm} = \frac{mv + 0}{m + m}$

$v_{cm} = \frac{v}{2}$

now the displacement of COM in above time interval is given as

$d = v_{cm} t$

$d = \frac{v}{2}\sqrt{\frac{2l}{g}}$

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Topic : center of mass

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