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प्रश्न
Consider a gravity-free hall in which an experimenter of mass 50 kg is resting on a 5 kg pillow, 8 ft above the floor of the hall. He pushes the pillow down so that it starts falling at a speed of 8 ft/s. The pillow makes a perfectly elastic collision with the floor, rebounds and reaches the experimenter's head. Find the time elapsed in the process.
उत्तर
Given,
Mass of the man, Mm = 50 kg
Mass of the pillow, MP = 5 kg
Velocity of pillow w.r.t. man, \[\vec{V}_{pm}\]= 8 ft/s
Let the velocity of man be \[\vec{V}_m\] .
\[\vec{V}_{pm} = \vec{V}_p - ( - \vec{V}_m )\]
\[ = \vec{V}_p + \vec{V}_m \]
\[ \Rightarrow \vec{V}_p = \vec{V}_{pm} - \vec{V}_m\]
As no external force acts on the system, the acceleration of centre of mass is zero and the velocity of centre of mass remains constant.
Using the law of conservation of linear momentum:
Mm × Vm = Mp × Vp ...(1)
\[M_m \times V_m = M_p \times ( V_{pm} - V_m )\]
\[ \Rightarrow 50 \times V_m = 5 \times (8 - V_m )\]
\[ \Rightarrow V_m = \frac{8}{11} = 0 . 727 \text{ ft/s }\]
∴ Velocity of pillow, \[\vec{V}_p\] = 8 − 0.727 = 7.2 ft/s
The time taken to reach the floor is given by,
\[t = \frac{s}{v} = \frac{8}{7 . 2} = 1 . 1 \text{ s}\]
As the mass of wall is much greater than the mass of the pillow,
velocity of block before the collision = velocity after the collision
⇒ time of ascent = 1.11 s
Hence, total time taken = 1.11 + 1.11 = 2.22 s
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