Question.
How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm and does not rebound ? Take its downward acceleration to be 10 ms–2.
How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm and does not rebound ? Take its downward acceleration to be 10 ms–2.
Solution:
Mass of dumb-bell, $m=10 \mathrm{~kg} ;$ initial velocity, $u=0 ;$ final velocity, $v=? ;$ distance,
$\mathrm{s}=80 \mathrm{~cm}=0.8 \mathrm{~m} ;$ acceleration, $\mathrm{a}=10 \mathrm{~ms}^{-2}$
We know
$\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}$
$\mathrm{v}^{2}-(0)^{2}=2 \times 10 \times 0.8$
$v^{2}=16$
$v=\sqrt{16}=4 \mathrm{~ms}^{-1}$
$\therefore$ Momentum of dumb-bell transferred to ground $=\mathrm{mv}=10 \times 4=40 \mathrm{~kg} \mathrm{~ms}^{-1}$.
Mass of dumb-bell, $m=10 \mathrm{~kg} ;$ initial velocity, $u=0 ;$ final velocity, $v=? ;$ distance,
$\mathrm{s}=80 \mathrm{~cm}=0.8 \mathrm{~m} ;$ acceleration, $\mathrm{a}=10 \mathrm{~ms}^{-2}$
We know
$\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}$
$\mathrm{v}^{2}-(0)^{2}=2 \times 10 \times 0.8$
$v^{2}=16$
$v=\sqrt{16}=4 \mathrm{~ms}^{-1}$
$\therefore$ Momentum of dumb-bell transferred to ground $=\mathrm{mv}=10 \times 4=40 \mathrm{~kg} \mathrm{~ms}^{-1}$.
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