Question

A glass ball whose mass is 10 g falls from a height of 40 m and rebounds to a height of 10m. Find the impulse and average force between the glass ball and the floor if the time during which they are in contact is 0.1 s

Answer 1

$<b><i>$

Given,
m = 10 g = $10×10^{-3}$ kg
s = 40 m
h = 10 m

Case-(1):
--- Find velocity of falling ball

We have,
$v^{2}-u^{2}=2as$
As it is falling freely u = 0
a = g = 9.8 m/s

=> $v = \sqrt{2gs}$
=> $v = \sqrt{2×9.8×40}$
=> $v = 28\: m/s$

--- Find Momentum just before hitting the ground

Say Downward Momentum as $P_{1}$

We have, P= m×v

$P_{1}= (10×10^{-3})× 28$

=> $P_{1}$ = 0.28 kgm/S = 0.28 N/s

Case-(2):
--- Find Velocity with which the ball rebounds(to height of 10m) after hitting the ground.

We have,
$v^{2}-u^{2}=2as$
As it is traveling upward v = 0
a = -g = -9.8 m/s
s = h = 10
Here distance refer to rebound height

=> $0- u^{2} = -2×9.8×10$
=> $u^{2} = 196$
=> $u = 14\: m/s$

--- Find Momentum just after hitting(or leaving) the ground

Say upward Momentum as $P_{2}$

We have, P= m×v

$P_{2}= (10×10^{-3})× 14$

=> $P_{2}$ = 0.14 kgm/S = 0.14 N/s

---------- Find 'F' using Change in Momentum

∆P = $P_{2} - P{1}$
F×t = $(0.14)_{upward} +(0.28)_{downward}$

Given
They are in contact in 0.1 sec
=> t = 0.1 sec

So

=> F×(0.1) = 0.14 + 0.28
=> 0.1 F = 0.42
=> F = 4.2 N

The Required impulse and Average Force between the glass ball and the floor is $4.2\: Newtons$

✌✌✌

Keep it Mello

Answer 2

Answer:

u = velocity before touching the ground

u^2 = 2gh

u^2 = 2×9.8×40

u^2 = 784

u = 28m/s

height of rebound = 10m

v = velocity sfter touching the ground

v^2 = 2gh

v^2 = 2×9.8×10

v^2 = 196

v = 14m/s

Therefore, impulse = mv-mu

= m(v-u)

= 100/1000[14-(-28)]

= 4.2Ns

Average force during the impact = 4.2Ns/0.1s

=42N