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8. FORCE AND LAWS OF MOTION
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How much momentum will a dumb-bell of mass $10\, kg$ transfer to the floor if it falls from a height of $80\, cm$ ? Take its downward acceleration to be $10\, m \,s^{-2}.$
A
$25\, kg\, m \,s^{-1}$
B
$49\, kg\, m \,s^{-1}$
C
$40\, kg\, m \,s^{-1}$
D
$45\, kg\, m \,s^{-1}$
Solution
Mass of the dumbbell, $m = 10\, kg$
Distance covered by the dumbbell, $s = 80\, cm = 0.8\, m$
Acceleration in the downward direction, $a = 10 \,m/s^2$
Initial velocity of the dumbbell, $u = 0$
Final velocity of the dumbbell (when it was about to hit the floor) $= v$
According to the third equation of motion:
$v^2= u^2 + 2as$
$v^2= 0 + 2 (10) 0.8$
$v= 4\, m/s$
Hence, the momentum with which the dumbbell hits the floor is $= mv $
$= 10 \times 4 \,kg \,m \,s^{-1} $
$= 40\, kg\, m \,s^{-1}$
Standard 9
Science
