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Centre of Mass, Linear Momentum, Collision — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsCentre of Mass, Linear Momentum, Collision5 Marks
Question
A man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling vertically. When at a height h from the ground, he notices that the ground below him is pretty hard, but there is a pond at a horizontal distance x from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just succeeds to avoid the hard ground, where will the bag land?
✓

Answer

Mass of man = M, Initial velocity = 0 Mass of bad = m Let the throws the bag towards left with a velocity v towards left. So, there is no external force in the horizontal direction. The momentum will be conserved. Let he goes right with a velocity$\text{mv}=\text{MV}$
$\Rightarrow\text{V}=\frac{\text{mv}}{\text{M}}$
$\Rightarrow\text{V}=\frac{\text{MV}}{\text{m}}\dots(\text{i})$
Let the total time he will take to reach ground $=\sqrt{\frac{\text{2H}}{\text{g}}}=\text{t}_1$ Let the total time he will take to reach the height h $=\text{t}_2=\sqrt{\frac{2(\text{H}-\text{h})}{\text{g}}}$ Then the time of his flying $=\text{t}_1-\text{t}_2=\sqrt{\frac{\text{2H}}{\text{g}}}-\sqrt{\frac{2(\text{H}-\text{h})}{\text{g}}}=\sqrt{\frac{2}{\text{g}\big(\sqrt{\text{H}}-\sqrt{\text{H}-\text{h}}\big)}}$ Within this time he reaches the ground in the pond covering a horizontal distance x$\Rightarrow\text{x}=\text{V}\times\text{t}\Rightarrow\text{V}=\frac{\text{x}}{\text{t}}$
$\therefore\text{v}=\frac{\text{M}}{\text{m}}\frac{\text{x}}{\text{t}}=\frac{\text{M}}{\text{m}}\times\frac{\sqrt{\text{g}}}{\sqrt2\big(\sqrt{\text{H}}-\sqrt{\text{H}-\text{h}}\big)}$
As there is no external force in horizontal direction, the x-coordinate of CM will remain at that position.$\Rightarrow0=\frac{\text{M}\times(\text{x})+\text{m}\times\text{x}_1}{\text{M}+\text{m}}$
$\Rightarrow\text{x}_1=-\frac{\text{M}}{\text{m}}\text{x}$
$\therefore$ The bag will reach the bottom at a distance (M/m) x towards left of the line it falls.

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