Question

Please solve the question in the attachment!​

Answer 1

Answer:

Question :-

☯ A ball of mass 0.15 kg is dropped from a height 10m strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly : ?

Given :-

☯ A ball of mass 0.15 kg is dropped from a height 10m strikes the ground and rebounds to the same height.

Find Out

☯ What is the magnitude of impulse imparted to the ball.

Solution :-

✭ In case of velocity of Ball at first :-

As we know that :

✯ v² = u² + 2as ✯

We have ;

• u = 0 m/s ,
• a = g = 10 m/s
• s = 10 m

So, according to the question or ATQ :-

➙ v₁² = (0)² + (2 × 10 × 10 )

➙ v₁² = (0 × 0) + (2 × 10 × 10 )

➙ v₁² = 2 × 10 × 10

➙ v₁² = 2 × 100

➙ v₁² = 200

➙ v₁ = 10√2 m/s

Here,

But the ball is dropped so the velocity would be negative and that is v₁ = - 10√2 m/s.

✭ In case of ball rebounds. :-

➙ Velocity of ball needed to reach the height is 10m.

➙ v₂ = √(2gh)

➙ v₂ = 10√2 m/s

Now, as we know that :

✯ Impulse = Change in linear momentum ✯

✯ Impulse = Final linear momentum - Initial linear momentum ✯

So, according to the question or ATQ :-

➙ Impulse = mv₁ - mv₂

➙ Impulse = (0.15 × 10√2) - (0.15 × -10√2)

➙ Impulse = 0.15 × 20√2

➙ Impulse ≈ 4.2

Henceforth, the magnitude of impulse imparted to the ball is nearly 4.2 kg m/s.

Correct options is (1) 4.2 kg m/s.

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Answer 2

Answer:

Question :-

A ball of mass 0.15 kg is dropped from a height 10m strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly : ?

Given :-

A ball of mass 0.15 kg is dropped from a height 10m strikes the ground and rebounds to the same height.

Find Out

What is the magnitude of impulse imparted to the ball.

Solution :-

✭ In case of velocity of Ball at first :-

As we know that :

✯ v² = u² + 2as ✯

We have ;

u = 0 m/s ,

a = g = 10 m/s

s = 10 m

So, according to the question or ATQ :-

➙ v₁² = (0)² + (2 × 10 × 10 )

➙ v₁² = (0 × 0) + (2 × 10 × 10 )

➙ v₁² = 2 × 10 × 10

➙ v₁² = 2 × 100

➙ v₁² = 200

➙ v₁ = 10√2 m/s

Here,

But the ball is dropped so the velocity would be negative and that is v₁ = - 10√2 m/s.

✭ In case of ball rebounds. :-

➙ Velocity of ball needed to reach the height is 10m.

➙ v₂ = √(2gh)

➙ v₂ = 10√2 m/s

Now, as we know that :

✯ Impulse = Change in linear momentum ✯

✯ Impulse = Final linear momentum - Initial linear momentum ✯

So, according to the question or ATQ :-

➙ Impulse = mv₁ - mv₂

➙ Impulse = (0.15 × 10√2) - (0.15 × -10√2)

➙ Impulse = 0.15 × 20√2

➙ Impulse ≈ 4.2

Henceforth, the magnitude of impulse imparted to the ball is nearly 4.2 kg m/s.

Correct options is (1) 4.2 kg m/s.

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