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Newton's laws for SQA Higher Physics

Newton's laws

This page covers the following topics:

1. Newton's first law
2. Newton's second law
3. Newton's third law
4. Newton's law of gravitation

Newton's first law states that an object will remain at rest or continue moving at a constant velocity, unless it is acted upon by an external resultant force. When there is no resultant force acting on an object, the object is said to be in equilibrium.

Newton's first law

Newton's second law states that the acceleration of an object is directly proportional to the force acting on it, in the direction of the force, and inversely proportional to the mass of the object. This means an object accelerates in the direction of the resultant force being exerted on it, the value of this acceleration depending on the mass of the object.

Newton's second law

Newton's third law states that when object A exerts a force on object B, then object B exerts an opposite force on object A. This implies that all forces exist in pairs which are called Newtonโ€™s third law pairs. These two forces must act on different objects, opposite directions, be of the same type of force and act along the same line and for the same time.

Newton's third law

A pull force that is exerted by two or more objects due to them having mass is called gravitational force. By Newton's third law, the force exerted by mass on the body is equal to the force exerted by the body on the mass. Newton's law of gravitation states that gravitational force is proportional to the product of the masses of the objects and inversely proportional to the square of distance between the objects.

Newton's law of gravitation

1

A spacecraft is between Earth and the Moon. The distance between Earth and the moon is 384400 km, the mass of the Earth is 6 ร— 10ยฒโด kg and the mass of the Moon is 7.35 ร— 10ยฒยฒ kg. Given that the resultant horizontal force on the star is 0, find the distance of the star from the centre of the Earth. Ignore gravitational forces from objects not mentioned in this question.

The horizontal forces acting on the spacecraft are the gravitational force of attraction from the Earth and the gravitational force of attraction from the Moon. Since the resultant force is 0, the two gravitational forces are equal.

F = GmM/rยฒ
(G ร— mass of Earth ร— mass of star)/dยฒ = (G ร— mass of Moon ร— mass of star)/(384400 ร— 10ยณ โˆ’ d)ยฒ

The mass of the spacecraft and the gravitational constant can be cancelled out.
Rearranging and simplifying gives 0.98775dยฒ โˆ’ 2 ร— 384400 ร— 10ยณ ร— d + (384400 ร— 10ยณ)ยฒ.
Solving for d gives 3.49 ร— 10โธ m (3 s. f.)

3.49 ร— 10โธ m

A spacecraft is between Earth and the Moon. The distance between Earth and the moon is 384400 km, the mass of the Earth is 6 ร— 10ยฒโด kg and the mass of the Moon is 7.35 ร— 10ยฒยฒ kg. Given that the resultant horizontal force on the star is 0, find the distance of the star from the centre of the Earth. Ignore gravitational forces from objects not mentioned in this question.

2

A rocket of mass 4 ร— 10โถ kg takes off vertically with a thrust of 6.2 ร— 10โท N. Calculate the initial acceleration of the rocket by using g = 9.8 m/sยฒ.

F = m ร— a
6.2 ร— 10โท N โˆ’ 4 ร— 10โถ ร— 9.8 = 4 ร— 10โถ kg ร— acceleration
acceleration = 5.7 m/sยฒ

5.7 m/sยฒ

A rocket of mass 4 ร— 10โถ kg takes off vertically with a thrust of 6.2 ร— 10โท N. Calculate the initial acceleration of the rocket by using g = 9.8 m/sยฒ.

3

A ping-pong ball of mass 0.2 kg is released from the bottom of a tube filled with water. Given that it accelerates towards the top at 1.6 m/sยฒ and that the upthrust is 62 N, find the drag acting on the ball.

resultant force = upthrust โˆ’ drag โˆ’ weight
ma = upthrust โˆ’ drag โˆ’ mg
0.2 kg ร— 1.6 m/sยฒ = 62 N โˆ’ 0.2 kg ร— 9.8 m/sยฒ โˆ’ drag
drag = 62 N โˆ’ 1.96 N โˆ’ 0.32 N = 59.7 N (3 s. f.)

59.7 N

A ping-pong ball of mass 0.2 kg is released from the bottom of a tube filled with water. Given that it accelerates towards the top at 1.6 m/sยฒ and that the upthrust is 62 N, find the drag acting on the ball.

4

Water is released out of a nozzle with a force of 2875 N from the back of a jet ski of mass 350 kg. Calculate the initial acceleration of the jet ski.

By Newton's third law, since the nozzle is exerting a backwards force on the water, the water exerts an equal and opposite force on the nozzle, and thus the jet ski, in the forward direction. An external resultant force is now acting on the jet ski, therefore by Newton's second law, it will accelerate in its direction.

F = m ร— a
2875 N = 350 kg ร— acceleration
acceleration = 2875 N รท 350 kg = 8.21 m/sยฒ

8.21 m/sยฒ

Water is released out of a nozzle with a force of 2875 N from the back of a jet ski of mass 350 kg. Calculate the initial acceleration of the jet ski.

5

A toy boat is powered by a small electric fan. Explain, using Newton's laws, how the toy moves forward.

When the fan is turned on, the fan exerts a force on the air in the backwards direction. By Newton's third law, the air exerts a force on the fan in the opposite direction, ie. the forward direction. This creates a resultant force on the boat in the forward direction, and so by Newton's second law, the boat accelerates in the direction of this force, and so moves forward.

The fan exerts a force on the air and an equal and opposite force is exerted on the fan. This generates a resultant force on the boat.

A toy boat is powered by a small electric fan. Explain, using Newton's laws, how the toy moves forward.

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