What is the y-intercept?

We have seen in a previous post the gradient of a straight line graph. Today we are going so have a look at the y-intercept.

Now as you know a graph is composed of two axes. The y-axis and the x-axis as shown in fig 1.

Fig 1

Now in order to find the y-intercept you will need a straight line that passes through the y-axis, i.e it intercept the y-axis,  as seen in fig 2 below.

Fig 2

As you can see in fig 2 above all three lines crosses the y-axis. As a result the three lines would have a y-intercept.

How to obtain the y-intercept of a straight line?

1. If the x-axis starts at 0

As you can see in fig 3 below the point at which the line crosses the y-axis is at the y=2 coordinate. Hence the y-intercept is 2.

We can thus define the y-intercept as being the value of the y-coordinate when the x-coordinate is 0. 

Fig 3

2. If the x-axis does not start at 0

As you can see in fig 4 below the x-axis does not start at 0. Hence as you may have guess wrongly the y-intercept is not 2 since according to the definition the y-intercept is the y-coordinate when the x-coordinate is 0.

So how do you obtain the y coordinate.

Fig 4

You will first have to calculate the gradient of the line using the method described in this post.

The gradient in this case is 1.

You will use the equation y = mx + c  and a coordinate on the line in this case (3,2).

y = 2

x=2

gradient = m =1

Hence the only variable left is c, the y-intercept.

3 = 1*2 + c

c = 3 – 2 =1

The y-intercept of the line is thus 1.

You should thus be very careful to check that the x-axis starts with 0 or does not start with 0 so as to choose which of the two methods to use.

Conservation of linear momentum in collisions

As we have seen in this previous post, in a close system the sum of linear momentum is a constant. Hence in any collision if the sum of linear momentum of the objects before the collision would be equal to the sum of linear momentum after the collision.

Fig 1 below shows two spherical objects travelling toward each other. The mass of the two objects are 1.0 kg each. Object 1 is travelling at a velocity of 5.0 m/s  and object 2 is travelling at a velocity of –5.0 m/s.

Fig 1

If  p_1b is the linear momentum of object 1 before the collision and p_2b

is the linear momentum of  object 2 before the collision then

the sum of linear momentum = p_1b +p_2b

= m₁u₁ + m₂v₁

=1.0*5.0 + 1.0*(-5.0)

= 5.0 - 5.0

=0 Kg m/s

As you can see before the collision the sum of the linear momentum is 0 kg m/s. According to the principle of conservation of linear momentum the sum of linear momentum at all time will thus be 0 kg m/s before and after the collision.

Fig 2 below shows the spherical objects after the collision while they are separating. According to the principle of conservation of linear momentum after the collision the sum of linear will thus be 0 kg m/s.

Question

If after the collision the two objects separate such that object 2 moves with a velocity of  5.0 m/s, calculate the velocity of object 1.

Fig 2

If  p_1a is the linear momentum of object 1 after the collision and p_2a

is the linear momentum of  object 2 after the collision then

the sum of linear momentum = p_1a +p_2a

= m₁u₂ + m₂v₂

=1.0*u₂  + 1.0 *5.0

= u₂  +5.0

Now if we use the principle of  of conservation of linear momentum

sum of linear momentum before the collision = sum of linear momentum before the collision

0 =   u₂  +5.0

u₂  = - 5.0 m/s

Newton’s second law of motion

We have seen in a previous post the Newton’s first law of motion.

Today we are going to see the second law of motion.

From our discussion in the first law of motion and in the post on acceleration when a force acts on an object either velocity increases due to a change in speed or the velocity changes due to a change in direction. Remember velocity is a vector quantity hence it will change if either the direction or the magnitude changes.

Hence we can deduce that if a force acts on an object then the velocity of the object changes.The more force is applied the greater the change in velocity. This must mean that if the mass remain constant then the momentum of the object changes. And hence the more force is applied the more the linear momentum will change.

Let us look at the situation below:

A force F acts on an object of mass m for a time t such that before the application of the force the object has a velocity v₁ hence a linear momentum of p₁ and after the force is applied the velocity of the object is v₂ hence the linear momentum is p₂.

Before the force is applied velocity = v₁ and linear momentum = p₁

After the force is applied velocity =va₂  and linear momentum is p₂.

Now the Newton’s second law of motion states that the force applied is directly proportional to the rate of change of linear momentum of the object.

If = a then

If k = 1

Then this will reduce the equation to

F= ma

This where the famous equation F = ma come from. This why it is also considered as the Newton’s second law of motion

Linear momentum and collisions

Linear momentum

What is linear momentum?

Principle of conservation of linear momentum

What is a collision?

conservation of linear momentum in collisions

Conservation of kinetic energy in collisions

Principle of conservation of linear momentum

The linear momentum is the product of the mass of an object and the velocity of an object.

Now if you have two objects that are moving toward each other or one object explodes into two different objects at all time the sum of the linear momentum of the objects will give a constant number.

Hence the principle of conservation of linear momentum states that in a close system the sum of linear momentum is constant.

The equation will be that at all time

p1 +p2+p3+p4+p5+ ………= constant

Consider the diagram below:

Fig 1 Before firing ball

Fig 2 After firing ball

As you can see from the diagram above, you have a close system that is made up of the the cannon and the cannon ball. Hence at all time the sum of the linear momentum will be a constant. Before the firing of the ball and after the firing of the ball.

Example

A canon ball is fired such that the ball leaves the cannon at a velocity of  150 m/s. If the cannon has a mass of 1000 kg and the cannon ball has a mass of 15 kg, determine the velocity of recoil of the cannon.

Hence if you apply the principle of conservation of linear momentum.

Before firing cannon ball

Sum of linear momentum before firing = p_ic  +p_ib

where p_ic is the initial momentum of the cannon and p_ib is the initial momentum of the ball.

After firing the cannon ball

sum of linear momentum after firing= p_fc  +p_fb

where p_fc is the final momentum of the cannon and p_fb is the final momentum of the cannon ball.

We are now going to determine the speed of recoil of the cannon after recoil.

Sum of linear momentum before firing = p_ic  +p_ib

= 1000*0 + 15*0

= 0 kg m/s

sum of linear momentum after firing= p_fc  +p_fb

=1000*v_rc +15*150

=1000*v_rc +2250

where is the v_rc velocity or recoil of the cannon.

Hence if the sum of linear momentum is a constant

then the sum of linear momentum before firing = sum of linear momentum after firing

0 = 1000*v_rc +2250

0 – 2250 = 1000*v_rc

-2250 = 1000*v_rc

v_rc = –2.25 m/s

This principles is used also when collisions are considered. We shall see collisions in a future post.

What is linear momentum?

Linear momentum is a physical quantity that depends on the mass and the velocity of the object.

Fig 1 below shows an object that has mass m and is moving with a velocity v.

 

The linear momentum which is denoted by the symbol p and is calculated using the equation below

 

linear momentum  = mass of object  x velocity of object

p = mv

Hence the unit for linear momentum is kg m s^-1

The linear momentum is thus a vector quantity since it is the product of a scalar quantity (the mass) and the vector quantity (the velocity).

We can thus say that the linear momentum of a body is defined as the product of the mass of that body and its velocity