47. A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be:

47. A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be:

(a) 9.9 m

(b) 10.1 m

(c) 10 m

(d) 20 m

Answer :  (b) 10.1 m

\( \text{Conservation of Momentum method} \)

\( m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \)

\( m_1(0) + m_2(0) = 50 v_1' + (0.5)(-2) \)

\( v_1' = \)\( \frac{1 \times 2}{50 \times 2} \)\( \)

\( = \)\( \frac{2}{100} \)\( \)

\( = 0.02 \, \text{m/s} \)

\( s = ut + \)\( \frac{1}{2} \)\( at^2 \)

\( 10 = 2t \)

\( t = 5 \, \text{s} \)

\( d = vt \)

\( = 0.02 \times 5 \)

\( = 0.1 \, \text{m} \)

distance of the man above the floor = 10 + 0.1 = 10.1 m

\( \text{Center of Mass method} \)

\( \text{Center of Mass (cm) does not change as there is no external force acting on system} \)

\( M R_{cm} = m_1 r_1 + m_2 r_2 \)

\( (50 + 0.5) \times 10 = 50 r_1 + 0.5(0) \)

\( r_1 = \)\( \frac{50.5 \times 10}{50} \)\( \)

\( = 10.1 \, \text{m} \)