Collisions and Momentum Conservation

​Newton's Laws of Motion

Subject | Subject

Some text here about the topic of discussion

• First and Second Law provide a general description of what happens to a body of mass m and its trajectory when an outside force F influences them.

• Third Law deals with an interaction between two bodies.

Collisions and Momentum Conservation

Poll

How many of you have played billiard (pool)?

• Yes, I played it.

• No, I haven't played it.

• No, I haven't played it but I watched others.

How to hit Ball 3 with Ball 1, without touching Ball 2?

​A Billiard Problem

Multiple Choice

In which direction will Ball 1 hit back from the edge?

Why did you choose this particular direction?

Multiple Choice

How will Ball 1 and Ball 2 move after their collision? Both balls have the same size and same mass.

Elaborate the choice of your answer.

• Both balls continue moving together to the left, at velocity lower than v1.

• Ball 1 reflects backward with the same velocity v1, Ball 2 keeps standing.

• Ball 1 stops, Ball 2 starts moving with the same velocity v1.

• Ball 1 knocks Ball 2 and reflects backward. Each ball travels with velocities lower than v1. (v1 = 2 x v2)

Before collision

v₁ = const.

v₂ = 0

The velocity hasn't changed in value or direction, but different bodies are moving before and after the collision.

Writing it down...

After collision

v₁' = 0

v₂' = v₁

Multiple Choice

Two balls of same size and mass (0.5 kg) move towards each other with velocities

v₁ = 1 m/s and v₂ = -0.5 m/s.

What are going to be their velocities after the collision? Elaborate your answer briefly.

• v1 = -1 m/s

  v2 = 0.5 m/s

• v1 = 0.5 m/s

  v2 = -1 m/s

• v1 = -0.5 m/s

  v2 = 1 m/s

• v1 = -0.75 m/s

  v2 = 0.75 m/s

Before collision

v₁ = 1 m/s

v₂ = -0.5 m/s

Minus (-) signifies the object is moving in direction opposite of our initial observation, asterisk (') denotes velocities after collision/event.

Writing it down... with numbers.

After collision

v₁' = -0.5 m/s

v₂' = 1 m/s

Ball 1

Δv₁ = v₁' - v₁

= -0.5 - 1 = -1.5 m/s

-Δv₁ = Δv₂

-(v₁' - v₁) = v₂' - v₂

v₁ + v₂ = v₁' + v₂'

Change in velocity

Ball 2

Δv₂ = v₂' - v₂

= 1 - (-0.5) = +1.5 m/s

​However, something is still missing.

Let's see what Sir Isaac Newton has to say!

m₁Δv₁ = -m₂Δv₂

m₁(v₁' - v₁) = m₂(v₂' - v₂)

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Third Newton's Law - some math

And if m₁ = m₂, we get the equality from previous problem,

i.e. v₁ + v₁' = v₂ + v₂'

​Conservation of momentum in 1D

​New physical quantity - momentum

Momentum p is a quantity of motion, expressed in kg m/s. Vector quantity derived from the product of object's mass and velocity.

​In a system not influenced by external forces (i.e. a closed system), it is a conserved quantity.

Multiple Choice

Car of mass 1000 kg at velocity of 50 m/s collides a bus of mass 5000 kg moving at velocity 5 m/s. Immediately after collision, both vehicles move together in same direction. Determine their velocity during that time.

• 12.5 m/s

• -12.5 m/s

• 4.17 m/s

• -4.17 m/s

m₁ = 1000 kg

v₁ = -50 m/s

--------------

V' = ?

m₁v₁ + m₂v₂ = (m₁ + m₂)V'

1000 ⋅ (-50) + 5000 ⋅ 5 = (1000 + 5000)V'

V' = -25000/6000 = -4.166 m/s ≈ -4.17 m/s

Car/bus crash solution

m₂ = 5000 kg

v₂ = 5 m/s

Poll

You are riding on a skateboard. In which direction, with respect to skateboard, you have to jump off in order to achieve the maximum possible speed of the skateboard?

• Sideways of skateboard.

• In front of the skateboard.

• Behind the skateboard.

Multiple Choice

Person, mass 50 kg, is riding on a skateboard with a velocity of 6 m/s. Person jumps of the skateboard with a velocity of 5 m/s. Which velocity will skateboard acquire if the person jumps off:

a) in direction of moving skateboard? v'

b) in direction opposite of moving skateboard? v''

• a) v´ = 7.2 m/s

  b) v´´ = 11 m/s

• a) v´ = 11 m/s

  b) v´´ = 61 m/s

• a) v´ = 7.2 m/s

  b) v´´ = 61 m/s

a)

(m_B + m_S)V = m_Bv_B' + m_Sv_S´

(50 + 10) ⋅ 6 = 50 ⋅ 5 + 10v_S´

v_S´ = 11 m/s

m_B = 50 kg, m_S = 10 kg, V = 6 m/s, v_B = 5 m/s

v_S´ = ?

Skateboarding solution

b)

(m_B + m_S)V = m_Bv_B' + m_Sv_S´

(50 + 10) ⋅ 6 = 50 ⋅ (-5) + 10v_S´

v_S´ = 61 m/s

Multiple Choice

Which of the following equations most accurately describes the collision as seen in the image on the left? Smaller sphere has mass 0.5 kg and initial velocity 1 m/s. Larger sphere has mass 1 kg.

Determine the velocities of spheres after the collision, v1' and v₂'.

• $$m_1v_1+m_2v_2=m_2v_2´$$

• $$m_1v_1=m_1v_1'\ +\ m_2v_2´$$

• $$m_1v_1=\left(m_1+m_2\right)v_1'$$

​Collisions solution

m₁ = 0.5 kg, m₂ = 1 kg

v₁ = 1 m/s, v₂ = 0 m/s

------------

v₁', v₂'

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

1 ⋅ 0.5 + 0 = 0.5v₁' + 1v₂'

• a problem: two unknowns in one equation

• one equation missing for the solution -> from conservation of energy

Vanja Blažinić

Thank you for your attention.