A rubber ball of mass 200 g falls from a height of 5 m and rebounds to
a height of 4m. Find the impulse and the average force between the ball
and the ground if the time during which they are in contact was 0.2 sec.
(Please Help)
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Explanation:
Given that,
Ball of mass = 50 g
Height = 1 m
Rebound to height = 0.5 m
Time = 0.1 s
Now,
Initial velocity of the ball =um/s
Let the final velocity =vm/s
Using equation of motion
v
2
=u
2
+2gh
v
2
=0+2×9.8×0.1
v=
1.96
v=1.4m/s
In the second time,
Final velocity=0m/s
Now, Using equation of motion
v
2
=u
2
−2gh
0=u
2
−2×9.8×0.5
u
2
=9.8
u=3.1m/s
Now, Impulse= change in linear momentum
=mv−m(−u)
=m(v+u)
=0.05×(1.4+3.1)
=0.225Ns
Now, the Force=Impulse/Time
F=
time
impulse
F=
0.1
0.225
F=2.25N
Hence, the impulse is 0.225 Ns and force is 2.25 N
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