A rubber ball of mass 100g falls from a height of 1 m rebounds to a height of 40cm find the impulse and the average force between the ball and the ground if time during which they are in contact was 0.1 s
Answers
Answer:
nitial velocity of the ball = u=0m/s
Let the final velocity = vm/s
g=9.8m/s²; h=10cm=0.1m
Using v²=u²+2gh we get,
v²=0+2×9.8×0.1=1.96
∴, v=√1.96=1.4m/s
In the second time,
Final velocity=0m/s
h=50cm=0.5m
Using, v²=u²-2gh [as the ball goes against gravitational acceleration]
0=u²-2×9.8×0.5
or, u=√9.8=3.13m/s
Now, Impulse= change in linear momentum
=mv-m(-u)
=0.05×(1.4+3.13)
=0.05×4.53
=0.2265Ns
∴, Force=Impulse/Time
=0.2265/0.1
=2.265N
Answer:
Impulse = 0.723 N/s
Average force = 7.23 N
Explanation:
When the ball is falling down:
u = 0
g = 9.8
h = 1
v² = u² + 2gh
v² = 0 + 2*9.8*1
v² = 19.6
v = √19.6 = 4.43
When the ball rebounds or goes up:
v = 0
g = 9.8
h = 40cm = 0.4m
v² = u² - 2gh
0 = u² - 2gh
u² = 2gh = 2*9.8*0.4 = 7.84
u = √7.84 = 2.8 m/s
Mass = 100g = 0.1kg
Impulse = mv - (-mu)
= m (v + u)
= 0.1 (4.43 + 2.8)
= 0.1 * 7.23
= 0.723 N/s
Average force = impulse/time = 0.723/0.1 = 7.23 N
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