## Position vs. Time and velocity vs. time for a free-falling object:

The position of a free-falling object is measured downward from a certain starting point.

The position of the object increases with time as it falls.

The graph would be a straight line with a negative slope, starting at the initial position and ending at the final position.

Velocity vs. Time graph for a free-falling object:

The velocity of a free-falling object increases with time as it falls.

The graph would be a straight line with a positive slope, starting at zero and ending at the final velocity.

To determine the acceleration due to gravity, we can use the equation:

a = v^2/s or a = 2s/t^2

Where a is the acceleration, v is the final velocity, s is the final position, and t is the time of fall.

Since the acceleration due to gravity is constant, the slope of the velocity vs time graph is the acceleration.

Position vs. Time:

The graph for position vs. time for a free-falling object would be a straight line that starts at a non-zero position (the initial height of the object) and decreases as time goes on. The slope of this line would represent the velocity of the object. The equation for this line would be:

y = -g*t + h

where g is the acceleration due to gravity, t is the time, and h is the initial height of the object.

Velocity vs. Time:

The graph for velocity vs. time for a free-falling object would be a straight line that starts at a non-zero velocity (the initial velocity of the object) and decreases as time goes on. The slope of this line would represent the acceleration of the object. The equation for this line would be:

y = -g*t + v

where g is the acceleration due to gravity, t is the time, and v is the initial velocity of the object.

Determining the Acceleration due to Gravity:

To determine the acceleration due to gravity, we can use the slope of either the position vs. time or velocity vs. time graph. For the position vs. time graph, the slope is -g, and for the velocity vs. time graph, the slope is -g. Therefore, we can conclude that the acceleration due to gravity is -g.

## Momentum

Momentum is a measure of an object’s inertia or its resistance to changes in its motion. It is calculated as the product of an object’s mass (m) and its velocity (v). So, the formula for momentum (p) is:

p = m x v

Where: p = momentum m = mass of the object (in kg) v = velocity of the object (in m/s)

Example: If an object has a mass of 5 kg and is moving at a velocity of 10 m/s, then its momentum would be: p = 5 kg x 10 m/s = 50 kg m/s

Momentum is a vector quantity, which means it has both magnitude and direction. The magnitude of momentum is simply the product of an object’s mass and velocity, as described by the formula p = mv. The direction of momentum is the same as the direction of an object’s velocity.

For example, consider a car moving east with a mass of 1000 kg and a velocity of 20 m/s. The magnitude of its momentum would be: p = mv = 1000 kg x 20 m/s = 20,000 kg m/s

The direction of the momentum would be in the east, the same as the direction of the car’s velocity.

Another example is a ball thrown upwards with a mass of 0.2 kg and a velocity of 10 m/s upward. The magnitude of its momentum would be: p = mv = 0.2 kg x 10 m/s = 2 kg m/s

The direction of the momentum would be upward, the same as the direction of the ball’s velocity.

It’s important to note that in collisions and other interactions between objects, the total momentum of a closed system (systems with no external forces) is conserved, which means that the momentum of all objects before and after the collision will be equal. This is the principle of conservation of momentum.

## Grade 11 Newton Laws QUESTION

A tow truck is towing a car using an inelastic steel cable (of negligible mass) as shown in the diagram below. The steel cable forms an angle of 30° with the horizontal.

The two vehicles move from rest on a straight, horizontal road to the right. The mass of the car is 1100 kg and the mass of the tow truck is 4000 kg. The tow truck’s engine applies a force of 15 000 N. A constant frictional force of 1456 N is acting on the car and a constant frictional force of 4520 N is acting on the tow truck respectively.

3.1 State Newton’s Second Law of motion, in words. (2)

3.2 Draw a labeled, free-body diagram of all the forces acting on the car. (4)

3.3 Calculate the:

3.3.1 acceleration of the car. (6)

3.3.2 magnitude of the tension T in the cable. (3)

3.5 Using equations of motion, calculate the distance that the car will travel in 4s. (2)

3.6 State Newton’s First Law of motion, in words. (2)

3.7 Use Newton’s laws of motion to explain why towing can be dangerous. (2)

3.8 If the force of horizontal tension in the cable, from the car on the truck, is 1890 N, what is the horizontal force of the truck on the car? Explain with reference to the relevant scientific principles. (2)