The mass of an iron ball is 90 kg which is dropped on a heap of sand fromm a height of 9.8 m so that it penitrates 2 cm inside it what resistive force is applied by the sand on the ball.
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Consider the motion of ball from A to B.
B→ just above the sand ( just to penetrate)
u=0,a=9.8 m/s2,s=5 m
The distance covered by the ball is:
S=ut+21at2
⇒5=0+21(9.8)t2
⇒t2=4.95=1.02
⇒t=1.01.
∴ velocity at B,v=u+at=9.8×1.01(u=0)=9.89 m/s.
From motion of ball in sand
u1=9.89 m/s,v1=0,a=?,s=10 cm=0.1 m
a=2sv12−u12=2×0.10−(9.89)2=−490 m/s2
The retardation in sand is 490 m/s2.
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