Question

A rubber ball of mass 100 gram falls from a height of 1 m rebounds to a height of 40 cm. Find the impulse and the average force between the ball and the ground if time during which they are in contact was 0. 1 sec

Answer 1

Given :-

Mass of the rubber ball = 100 g

Height = 1 m

Rebound to height = 40 cm

Time taken = 0.1 sec

To Find :-

The impulse and the average force between the ball and the ground if time during which they are in contact was 0.1 sec.

Analysis :-

Firstly find the final velocity using the third equation of motion. Using the same, you can easily find the initial velocity as well.

Then using the formula of change in linear momentum and find the impulse accordingly.

In order to find the force, divide the impulse by time.

Solution :-

We know that,

• v = Final velocity
• g = Gravity
• u = Initial velocity
• h = Height

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2=u^2+ 2gh}}$

Given that,

Initial velocity (u) = 0 m/s

Gravity (g) = 9.8 m/s

Height (h) = 1 m

Substituting their values,

⇒ v² = 0² + 2 × 9.8 × 1

⇒ v² = 19.6

⇒ v = √19.6

⇒ v = 4.4 m/s

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2=u^2- 2gh}}$

Given that,

Final velocity (v) = 0 m/s

Gravity (g) = 9.8 m/s

Height (h) = 40 cm = 0.4 m

Substituting their values,

⇒ 0² = u² - 2 × 9.8 × 0.4

⇒ u² = 7.8

⇒ u = √7.8

⇒ u = 2.7 m/s

Given that,

• p = Momentum
• m = Mass
• v = Final velocity
• u = Initial velocity

Using the formula,

$\underline{\boxed{\sf Impulse= m(v+u) }}$

Given that,

Mass (m) = 100 g = 0.1 kg

Final velocity (v) = 4.4 m/s

Initial velocity (u) = 2.7 m/s

Substituting their values,

⇒ p = 0.1 (4.4 + 2.7)

⇒ p = 0.1 × 7.1

⇒ p = 0.71 Ns

Using the formula,

$\underline{\boxed{\sf Force=\dfrac{Impulse}{Time} }}$

Given that,

Impulse = 0.71 Ns

Time (t) = 0.1 sec

Substituting their values,

⇒ f = 0.71/0.1

⇒ f = 7.1 N

Therefore, the average force is 7.1 N.

Answer 2

Answer:

Given :-

Mass of the rubber ball = 100 g

Height = 1 m

Rebound to height = 40 cm

Time taken = 0.1 sec

To Find :-

The impulse and the average force between the ball and the ground if time

Solution :-

Third Equation

$\sf \: {v}^{2} = {u}^{2} + 2gs$

v² = 0² + 2 × 9.8 × 1

v² = 2 × 9.8

v² = 19.6

v = √19.6

v = 4.4 m/s

$\sf \: s = {v}^{2} - u{}^{2}= 2gh$

0² = u² - 2 × 9.8 × 0.4

u² = 7.8

u = √7.8

u = 2.7 m/s

$\sf \: Impulse = m(v + u)$

p = 0.1 (4.4 + 2.7)

p = 0.1 × 7.1

p = 0.71 Ns

$\sf \: Force = \dfrac{Impulse}{ Time }$

Force = 0.71/0.1

F = 7.1 N