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**GIven data:
**mass = 4 kg

velocity = 10 m/s

momentum =?

**According to formula of momentum,
**p = mv

p = 4(10)

p = 40 kg m/s

**From the figure,
**gravitational force acting on skydiver = 800 N

air resistance force acting on skydiver = 800 N

mass of skydiver = 100 kg

So,

net force,

F = 800 – 800

F = 0 N

acceleration,

a = F / m

a = 0 / 100

a = 0 m/s2

**From the figure,
**gravitational force acting on skydiver = 800 N

air resistance force acting on skydiver = 600 N

mass of skydiver = 100 kg

So,

net force,

F = 800 – 600

F = 200 N

acceleration,

a = F / m

a = 200 / 100

a = 2 m/s2

**From the figure,
**gravitational force acting on skydiver = 800 N

air resistance force acting on skydiver = 250 N

mass of skydiver = 100 kg

So,

net force,

F = 800 – 250

F = 550 N

acceleration,

a = F / m

a = 550 / 100

a = 5.5 m/s2

**From the figure,
**gravitational force acting on skydiver = 800 N

air resistance force acting on skydiver = 0 N

mass of skydiver = 100 kg

So,

net force,

F = 800 N

acceleration,

a = F / m

a = 800 / 100

a = 8 m/s2

(In this case, Stone hits the ground first)

(Both feather and a stone strike the ground at the **same time**)

**Given data:
**initial velocity = 60 m/s

final velocity = 30 m/s

opposing force = – 4 N

mass = 400 grams = 0.4 kg

time = ?

**According to Newton’s second law formula,
**F = m (v – u) / t

(- 4) = 0.4 (30 – 60) / t

(- 4 × t) = 0.4 (- 30)

(- 4 × t) = – 12

4 t = 12

t = 12 / 4

t = 3 seconds

**Given data:
**mass = 400 kg

initial velocity = 0 m/s

after time = 20 seconds

final velocity = 25 m/s

net force =?

**According to Newton’s second law formula,
**F = m (v – u) / t

F = 400 (25 – 0) / 20

F = 10000 / 20

F = 500 N

**Given data:
**mass = 5 kg

initial velocity = 4 m/s

net force = (15 + 25) = 40 N

final velocity = ?

time = 10 seconds

**According to Newton’s second law formula,
**F = m (v – u) / t

40 = 5 (v – 4) / 10

5 (v – 4) = (40 × 10)

5 (v – 4) = 400

(v – 4) = 400 / 5

(v – 4) = 80

v = 80 + 4

v = 84 m/s

**Given data:
**mass = 5 kg

initial velocity = 35 m/s

opposing force = – 15 N

final velocity = ?

time = 4 seconds

**According to Newton’s second law formula,
**F = m (v – u) / t

(- 15) = 5 (v – 35) / 4

5 (v – 35) = (- 15) × (4)

5 (v – 35) = – 60

(v – 35) = (- 60) / 4

(v – 35) = – 12

v = – 12 + 35

v = 23 m/s

**Given data:
**mass = 1 kg

initial velocity = 0 m/s

net force = 5 N

final velocity = ?

time = 6 seconds

**According to Newton’s second law formula,
**F = m (v – u) / t

5 = 1 (v – 0) / 6

v = 5 × 6

v = 30 m/s

**Given data:
**mass = 100 kg

initial velocity = 20 m/s

final velocity = 50 m/s

time = 4 seconds

net force =?

**According to Newton’s second law formula,
**F = 100 (v – u) / t

F = 100 (50 – 20) / 4

F = 100 (30) / 4

F = 25 (30)

F = 750 N

**Given data:
**mass = 20 kg

momentum = 400 kg m/s

velocity =?

**According to formula of momentum,
**p = mv

v = p / m

v = 400 / 20

v = 20 m/s

**Given data:
**velocity = 2 m/s

momentum = 2000 kg m/s

mass =?

**According to formula of momentum,
**p = mv

m = p / v

m = 2000 / 2

m = 1000 kg

**Given data:
**mass = 5 kg

momentum = 25 m/s

velocity =?

**According to formula of momentum,
**p = mv

v = p / m

v = 25 / 5

v = 5 m/s

**Given data:
**mass = 200 grams = 0.2 kg

velocity = 4 m/s

momentum =?

**According to formula of momentum,
**p = mv

p = 0.2(4)

p = 0.8 kg m/s

**Given data:
**mass = 75 kg

momentum = 4500 kg m/s

velocity =?

**According to formula of momentum,
**p = mv

v = p / m

v = 4500 / 75

v = 60 m/s

(As it is clearly mentioned in the question that friction and air resistance is neglected)

Therefore,

According to the question, the car will continue to stay in motion with the same speed and in the same direction.

(As there is no such friction and air resistance to slow down the car)

**Always remember,
**An external force is

External force only helps in accelerating the object forward.

(Object will not change its behaviour if

According to the viewpoint of Loki, he believes that whatever forces acting on the book are in balanced condition.

So, there is no such motion of the book possible in the horizontal direction.

BUT,

Remember this flowchart discussed in Newton’s first law?

If Loki has understood Newton’s laws properly, then he knows that…

If the friction and air resistance are neglected, there is no such kind of air drag, right.

So, the object starts moving in the horizontal direction. (Even if the forces are balanced)

Therefore, Sara is correct.

**Always Remember,
**An external force is

External force only helps in accelerating the object forward.

(Object will not change its behaviour if

Well, option B is also correct.

But, as it is clearly mentioned in the question, what if no **external force** is given to the ball?

(In fact, it is also given that the ball is moving with the **uniform motion**)

If you remember the first law, it is clearly mentioned that…

An object will change its behaviour only if an unbalanced force acts on it. (Otherwise NOT)

**In short, remember this point:
**An external force is

External force only helps in accelerating the object forward.

(Object will not change its behaviour if

**Given data:
**mass = 1000 kg

net force = 4500 N

acceleration =?

**According to Newton’s second law,
**F = ma

a = F / m

a = 4500 / 1000

a = 4.5 m/s2

**Given data:
**acceleration = 4 m/s2

mass = 1.5 kg

net force =?

**According to Newton’s second law,
**F = ma

F = 1.5(4)

F = 6 N

**Given data:
**net force = 6 N

acceleration = 3 m/s2

mass =?

**According to Newton’s second law,
**F = ma

m = F / a

m = 6 / 3

m = 2 kg

**Given data:
**mass = 5 kg

acceleration = 4 m/s2

net force =?

**According to Newton’s second law,
**F = ma

F = 5(4)

F = 20 N

**Given data:
**mass = 5 kg

net force = 15 N

acceleration =?

**According to Newton’s second law,
**F = ma

a = F / m

a = 15 / 5

a = 3 m/s2

Given data:

acceleration = 3 m/s2

mass = 2 kg

net force =?

**According to Newton’s second law,
**F = ma

F = 2(3)

F = 6 N

When the brake is applied by the truck driver, the force will act in the backward direction.

(as the direction of acceleration will be similar to that of force)

When the truck is moving forward, the force will also act in the forward direction.

(as the direction of acceleration will be similar to that of force)

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