Question

7. A glass marble whose mass is 0.1 kg falls from a
height of 40 m and rebound to a height of 10 m. Find
the impulse and the average force between the
marble and the floor if the time during which they
are in contact is 0.1 s. (Take g = 9.8 ms-2)​

Answer 1

Answer :

• Impulse between the marble and the floor = 1.4 kg m/s
• Average force between the marble and the floor = 14 N

Explanation :

Given :

• Mass of the glass marble, m = 0.1 kg.
• Height from which the glass marble fall, h = 40 m.
• Height acquire by the glass marble after rebounding, h' = 10 m.
• Time of contact, t = 0.1 s.
• Initial velocity = 0 m/s

[Note : Initial Velocity of the glass marble will be zero, since the body is falling and rebounding with same velocity (Velocity with either falling or Velocity with rebounding)]

To find :

• Impulse between the marble and the floor, I = ?
• Average force between the marble and the floor, F = ?

Knowledge required :

• By definition, Impulse is the product of the mass of the body and change in velocity.i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀I = m∆v

Where :

⠀⠀⠀⠀⠀⠀⠀● I = Impulse

⠀⠀⠀⠀⠀⠀⠀● m = Mass

⠀⠀⠀⠀⠀⠀⠀● ∆v = Change in velocity

• By definition, Force is the product of the mass and acceleration of the body.i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀F = ∆p/∆t

Where :

⠀⠀⠀⠀⠀⠀⠀● F = Force

⠀⠀⠀⠀⠀⠀⠀● ∆p = Change in momentum

⠀⠀⠀⠀⠀⠀⠀● ∆t = change in time

We know that, Impulse is the change in momentum of a body , so we get the new Equation for force as,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀F = I/∆t

• Third equation of motion :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀v² = u² + 2gh

Where :

⠀⠀⠀⠀⠀⠀⠀● v = Final Velocity

⠀⠀⠀⠀⠀⠀⠀● u = Initial Velocity

⠀⠀⠀⠀⠀⠀⠀● g = Acceleration due to gravity

⠀⠀⠀⠀⠀⠀⠀● h = height

Solution :

Velocity of the body after reaching the ground :

Let the velocity be v'.

By using the third equation of motion and substituting the values in it, we get :

⠀=> v² = u² + 2gh

⠀=> v'² = 0² + 2gh

⠀=> v'² = 2gh

⠀=> v'² = 2 × 9.8 × 40

⠀=> v'² = 784

⠀=> v' = √784

⠀=> v' = 28

⠀⠀⠀⠀⠀⠀⠀∴ v' = 28 m/s

Hence the velocity with which the ball will reach the ground is 28 m/s.

Velocity of the body after rebounding from the ground :

Let the velocity be v.

By using the third equation of motion and substituting the values in it, we get :

⠀=> v² = u² + 2gh

⠀=> v² = 0² + 2gh

⠀=> v² = 2gh

⠀=> v² = 2 × 9.8 × 10

⠀=> v² = 196

⠀=> v = √196

⠀=> v = 14

⠀⠀⠀⠀⠀⠀⠀∴ v = 14 m/s

Hence the velocity with which the ball will rebound from the ground is 14 m/s.

Impulse imparted by the glass marble :

By using the equation for Impulse and substituting the values in it, we get :

⠀=> I = m∆v

But ∆v is the change in Velocity,i.e, (v2 - v1), by substituting it in the equation, we get :

⠀=> I = m(v' - v)

⠀=> I = 0.1 × (28 - 14)

⠀=> I = 0.1 × 14

⠀=> I = 1.4

⠀⠀⠀⠀⠀⠀∴ I = 1.4 kg m/s

Force exerted by the glass marble :

By using the first Equation of Motion and substituting the values in it, we get :

⠀=> F = I/∆t

⠀=> F = 1.4/0.1

⠀=> F = 14/10/1/10

⠀=> F = 14/10 × 10

⠀=> F = 14

⠀⠀⠀⠀⠀⠀⠀⠀∴ F = 14 N

Therefore,

• Impulse between the marble and the floor, I = 1 kg m/s
• Average force between the marble and the floor, F = 14 N.

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

A glass marble whose mass is 0.1 kg falls from a height of 40 m and rebound to a height of 10 m. Find the impulse and the average force between the marble and the floor if the time during which they

are in contact is 0.1 s. (Take g = 9.8 ms-2)

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• Time of contact = 0.1 second
• Mass of glass marble = 0.1 kg
• Initial velocity = 0 m/s
• Height from which the glass marble is falled = 40 metres
• Rebound to a height = 10 m

$\large{\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

• Average force be the marble nd the floor.
• Impulse between the marble nd the floor.

${\bigstar}\large{\boxed{\sf{\pink{Solution}}}}$

• Average force be the marble nd the floor = 1.4 kg m/s²
• Impulse between the marble nd the floor = 14N

${\bigstar}\large{\boxed{\sf{\pink{Full \: Solution}}}}$

Body's velocity after reaching the ground.

Let the velocity = v¹

⚪ Using 3rd equation of motion we have

to substitute the values well.

↝ v² = u² + 2gh

↝ v¹² = 0² + 2gh

↝ v¹² = 2gh

The value of g nd h are given so we have to apply it now :)

↝ v¹² = 2 × 9.8 × 40

↝ v¹² = 80 × 9.8

↝ v¹² = 784

↝ v¹ = √784

√ means Square root.

↝ v¹ = 28

$\large{\boxed{\boxed{\sf{28 m/s² \: is \: Answer}}}}$

Body's velocity rebounding from the ground.

Let the velocity = v

⚪Using 3rd equation of motion we have

to substitute the values well.

↝ v² = u² + 2gh

↝ v² = 0² + 2gh

↝ v² = 2gh

The value of g nd h are given so we have to apply it now :)

↝ v² = 2 × 9.8 × 10

↝ v² = 20 × 9.8

↝ v² = 196

↝ v = √196

√ means square root.

↝ v = 14

$\large{\boxed{\boxed{\sf{14 m/s² \: is \: Answer}}}}$

Impulse imparted by the glass marbel.

⚪Using 3rd equation of impulse we have

to substitute the values well.

↝ l = m∆v

But we know that ∆v means Change in velocity ( v² - v¹ )

Substitute the values we get :

↝ l = m(v¹ - v)

Value of m = given , Value of v¹ & v we already Finded.

↝ l = 0.1 × ( 28 - 14 )

↝ l = 0.1 × 14

↝ l = 1.4

$\large{\boxed{\boxed{\sf{1.4 kg m/s² \: is \: Answer}}}}$

Force exerted by glass marble

⚪Using 1st equation of motion we have

to substitute the values well.

↝ F = l/∆t

↝ F = 1.4 / 0.1

↝ F = 14/10/1/10

↝ F = 14/10 × 10

↝ F = 14

$\large{\boxed{\boxed{\sf{14 N \: is \: Answer}}}}$

$\huge{\boxed{\rm{\red{Explore \: more}}}}$

$\large\purple{\texttt{What is force ?}}$

The product of mass & body's acceleration is know to be $\large\purple{\texttt{force.}}$

$\large\blue{\texttt{Formula ?}}$

$\large{\boxed{\sf{f \: = \: p∆ / t∆}}}$

$\large\green{\texttt{f = p∆ / t∆ meaning}}$

$\mapsto$ F = force

$\mapsto$ ∆p = Change in momentum

$\mapsto$ ∆t = Change in time

$\large\purple{\texttt{What is impulse ?}}$

The product of mass of the body nd a change in velocity is known to be $\large\purple{\texttt{impulse.}}$

$\large\blue{\texttt{Formula ?}}$

$\large{\boxed{\sf{I \: = \: m∆v}}}$

$\large\green{\texttt{I = m∆v meaning}}$

$\mapsto$ I = Impulse

$\mapsto$ m = Mass

$\mapsto$ ∆v = Change in velocity

According to the both the definition nd formulas we combined something's as seen below ☞

$\qquadd$ F = I / ∆t

⚪Third eq of motion

$\qquadd$ v² = u² + 2gh

$\large\green{\texttt{v² = u² + 2gh}}$

$\mapsto$ v = Final velocity

$\mapsto$ g = Acceleration due to the gravitational force

$\mapsto$ u = Initial velocity

$\mapsto$ h = Height

@Itzbeautyqueen23

Hope it's helpful

Thank you :]