Forces, movement, shape and momentum (continued)

1.20 know and use the relationship between momentum, mass and velocity: 

momentum (kg m/s) = mass (kg) x velocity (m/s)
             p                   =     m         x        v

1.21 use the idea of momentum to explain safety features 

force = change in momentum/time taken

This means that if the time taken for the momentum to change is increased, then the overall force will be smaller (divided by bigger value for time.) So, increasing the time taken for a change in momentum will reduce the force felt, which can be helpful in creating safety features for vehicles. For example, crumple zones increase the time it takes for a cars momentum to reach zero (to have stopped,) and therefore decrease the force felt by the car and the people in it. 

1.22 use the conservation of mass to calculate the mass, velocity or momentum of objects.

From formula triangle, it is evident that: 

• mass = momentum/velocity 
• velocity = momentum/mass 
• momentum = mass x velocity 

1.23 use the relationship between force, change in momentum, and time taken: 

force = change in momentum/time taken 

1.24 demonstrate an understanding of Newton's third law 

Newton's third law states that: For every action, there is an equal and opposite reaction. So, whenever a body exerts a force on another, that force is being exerted back, but in the opposite direction. One example would be a car on the road, which is exerting a downward force on the ground (due to gravity,) but the floor is exerting an upwards force, hence why the car is not sinking into the ground. Another example is shown in the picture below: one skater is exerting a force on the other, and as a result vice versa.

1.25 know and use the relationship between the moment of a force and its distance from the pivot: 

moment (Nm) = force (N) x perpendicular distance from pivot (m)

For example, if the force applied to a point is 10N and its perpendicular distance from the pivot is 0.8m, then the moment = force x perpendicular distance from pivot

 = 10N x 0.8m 

 = 8Nm.

1.26 recall that the weight of a body acts through its centre of gravity  

An objects's centre of gravity is where all of its weight acts through (weight is directly related to the objects centre of gravity, weight = mass x gravitational field strength.) 

1.27 know and use the principal of  moments for a simple system of parallel forces acting in one plane 

The principal of moments states that when the sum of the clockwise moments is equal to the sum of the anticlockwise moments, then the object will be balanced. 

These moments are equal (20N), so the object is balanced. Here, the sum of anticlockwise moments = sum of clockwise moments. 

The weight and distance are equal on both sides, so the moments are too. Because the sum of anticlockwise moments = sum of clockwise moments, the see-saw is said to be balanced. 

1.28 understand that the upward forces on a light beam, supported at its ends, vary with the position of a heavy object placed on the beam 

This means that, say you had a beam being held in balance by two springs, and you added a force, ie 10N to the beam, the upward force of the beam would have to exert more force (10N) so they equal the downward force (become balanced again.)

1.29 describe experiments to investigate how extension varies with applied force for helical springs, metal wires and rubber bands

1. Attach a spring to a newton metre and measure it's length 
2. Add a specific  weight (ie 50g) and measure again
3. Contiue to add more weight, preferably going up in the same units (50g each time)
4. Do this until you have measured 6-8 different weights, so up until you have added 300g-400g.
5. Plot a graph from your results, with the weights on one axis, and the lengths on the other. From the graph, it should be evident that as the force increases, so does the length of the spring (it s extension.) 

1.30 understand that the initial linear region of a force-extension graph is associated with Hooke's law. 

Hooke's law states that: the extension of an elastic object is directly proportional to the force applied to it. The initial linear region (straight, diagonal line) of a force-extension graph would show that the extension of said object increases as more force is applied, this is Hooke's law. 

The initial linear region of the graph shows that the extension of the elastic object is directly proportional to the force applied to it (Hooke's law.) The curve in the graph represents the object having reached its elastic potential (will not return to original shape after force is removed.)

1.31  describe elastic behaviour as the ability of a material to recover its original shape after the forces causing deformation have been removed. 

The initial linear part of the force extension graph above shows the correlation between the extension of an elastic object and the force applied to it. The straight line only continues up until the point where the object can no longer recover its original shape after having the force placed upon it, removed. This means the object has reached its elastic potential (what its elastic behaviour is) and can no longer recover its original shape, so no longer performs its elastic behaviour.