A physics teacher: collisions

Showing posts with label collisions. Show all posts

Friday, February 26, 2010

Conservation of kinetic energy in collisions

As we have seen in this post during a collision between two object the velocity of the two colliding objects changes. Hence you would agree that he kinetic energy of the objects would change.

We have also seen in this post that in all collisions the sum of linear momentum is a constant. This is the principle of conservation of linear momentum. However as we are going to see in some collisions, elastic collisions, the sum of kinetic energy is also constant. That is the sum of the kinetic energies of the objects before the collision is the same as the sum of the kinetic energies after the collision.

We are now going to see an example of how to use this “principle” which i am going to call the “principle of conservation of kinetic energy”.

Fig 1 below shows two objects travelling towards each other and fig 2 shows the two objects separating after the two objects have separated.

So if the principle of conservation of kinetic energy applies then it means that Sum of kinetic energy before collision = sum of kinetic energy after collision

Fig 1 Before collision

Before the collision

Sum of kinetic energy before collision = kinetic energy of object 1 + kinetic energy of object 2

= ¹/₂m₁u₁² + ¹/₂m₂v₁²

Fig 2 After collision

After the collision

Sum of kinetic energy after collision = kinetic energy of object 1 + kinetic energy of object 2

= ¹/₂m₁u₂² + ¹/₂m₂v₂²

If you refer back to the principle then

Sum of kinetic energy before collision = sum of kinetic energy after collision

¹/₂m₁u₁² + ¹/₂m₂v₁²= ¹/₂m₁u₂² + ¹/₂m₂v₂²

Example

Two balls of mass 1.0 kg are travelling towards each other at a speed of 5.0 m/s towards the right (ball 1) and 5.0 m/s towards the left (ball 2) respectively. If after separation one of them moves moves off a speed of 5 m/s towards the right (ball 2 ) calculate the speed of the other ball.

Sum of kinetic energy before collision = sum of kinetic energy after collision

¹/₂m₁u₁² + ¹/₂m₂v₁²= ¹/₂m₁u₂² + ¹/₂m₂v₂²

¹/₂*1.0*5.0² + ¹/₂*1.0*(-5.0)²= ¹/₂*1.0*u₂² + ¹/₂*1.0*5.0²

Solving for u₂ the velocity of object 1 after the collision

12.5+ 12.5 = 0.5*u₂²+12.5

12.5= 0.5*u₂²

12.5 /0.5 = 25 =u₂²

u₂²= 25

There can be two solution to u₂

First u₂=-5 m/s and secondly u₂= 5 m/s

Since the second ball is already travelling towards the right and according to the principle of conservation of linear momentum then the only solution is

u₂=-5 m/s

Tuesday, February 23, 2010

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

Saturday, February 13, 2010

What is a collisions?

A collision is a phenomenon that can be described in three distinct steps.

1. Firstly the two objects that would collide must approach each other. The two objects may be travelling towards each other as shown below in fig 1. They may also be moving in the same direction as in Fig 2 but if the first one is slower than the second one they would meet at a certain time.

Fig 1 Two objects moving towards each other directly

Fig 2 Two objects moving in the same direction but about moving closer to each other.

2. The two objects that are travelling towards each other must meet. That is they make contact. During this contact they interact by exerting forces on each other. The forces that they exerts on each other must be equal to each other according to Newton’s third law of motion.

Fig 3 The two objects make contact and exert forces on each other

3. The two objects then separate and move away from each other in opposite as shown in fig 4 or moves in the same direction but with increasing distance of separation as shown in fig 5.