First slide

Second law

Question

Consider a large block placed on a smooth horizontal surface, with a man standing at one end of the block. The man walks to the other end, relative to the block. The distances (absolute) moved by the man and the block are:

Moderate

A

In the inverse ratio of their masses
B

In the direction ratio of their masses
C

Independent of their masses
D

Dependent both on their masses and speeds.

Solution

Let m and M be the respective masses of the person and the block, and let l be its length. Now, if $x_{1} and x_{2}$ be the absolute displacement of the person and the block and $x$ that of the centre of mass, then
$x = \frac{{mx}_{1} + {mx}_{2}}{m + M} \Rightarrow ∆ x = \frac{m ∆ x_{1} + m ∆ x_{2}}{m + M}$

Since horizontal forces are absent, so $∆ x = 0$
$m ∆ x_{1} + M ∆ x_{2} = 0 \Rightarrow m (- l) + (m + M) ∆ x_{2} = 0$
$∆ x_{2} = \frac{ml}{m + M}$
Similarly, $∆ x_{1} = \frac{Ml}{m + M}$

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