A ball of mass 1 kg is dropped from height 9.8 m, strikes with ground and rebounds at height of 4.9 m, if the
time of contact between ball and ground is 0.1 sec. then find impulse and average force acting on ball.
Answers
Explanation:
Velocity before strice = u downward
Velocity after strike = V upward
u=
2gS
1
=
2×9.8×9.8
=9.8
2
m/s g = 9.8 taking
V=
2gS
2
=
2×4.9×9.8
=9.8m/s
change in momentum = impulse
mu−(−mv)=F×t
F=
t
1×9.8
2
+9.8
=
t
9.8(
2
+1)
=
0.1
9.8(1.41+1)
=98×2.41=236.2N
Answer:
Explanation:
A ball of mass 1kg dropped from 9.8m height strikes the ground and rebounds to a height of 4.9m. If the time of contact between ball and ground is 0.1s, then find impulse and average force acting on ball.
Velocity before strice = u downward
Velocity after strike = V upward
u=2gS1 =2×9.8×9.8 =9.82 m/s g = 9.8 taking
V=2gS2 =2×4.9×9.8 =9.8m/s
change in momentum = impulse
mu−(−mv)=F×t
F=1×9.82 +9.8
---------------
t
=9.8(2 +1)
-----------
t
=0.1 9.8(1.41+1)
------------------ =98×2.41=236.2N
0.1
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