COLLISIONReporters:

Cabag Andaluz

Diaz Janier

CRASHESBILLIARDS

The act or process of colliding; a crash or conflict. A brief dynamic event consisting of the close

approach of two or more particles, such as atoms, resulting in an abrupt change of momentum or

exchange of energy.

MOMENTUM Involves motion-to have momentum, an object must be moving at a certain velocity Involves mass-the more massive, the more difficult to change your state immediately

-is the measure of one’s motion, equivalent to the product of one’s mass and velocity.

EXPRESS IN:

p= m v

Where p is the momentum ain kg-m/s

SAMPLE CONDITIONS:

1. If you have a VW Bug going 50 km/h and a truck going at the same velocity, which has more momentum?

2. What about two trucks, one moving at 25 km/h and the other moving 50 km/h. Which has more momentum?

3.

Example:

Calculate the momentum of 110 – kg missile traveling at 100 m/s eastward.

Given: Mass (m) – 110 kg of missileVelocity (v) – 100 m/s

Momentum - ?

Solution:

Direct substitution

p = mv = 110 kg (100 m/s) = 11000 kg – m/s, eastward

IMPULSE

Force (F) applied by one object to another object within a given time interval ∆t

Express in:F∆t= ∆pWhere F∆t is the impulse in newton-second

(N-s)

DERIVATION:

From Newton’s Law:F=m a

Since acceleration is:

F = m

Transporting ∆t to the other side of equation: F ∆t = m ∆v

F ∆t= ∆p

CONSERVATION OF MOMENTUM

All collisions have to conserve momentum.

Elastic Collision in One Dimension

=

since

( /

CASE 1: BOTH MASSES ARE EQUAL (

Stops

𝑣2′=𝑣1

CASE 2: MASS OF MOVING OBJECT GREATER THAN THE MASS OF THE STATIONARY OBJECT ()

for

EXAMPLE (CASE 1)

Before collision

After collision

EXAMPLE (CASE 2)

2 kg 1 kg

1 m/s = (1/3) m/s

1 m/s = (4/3) m/s

Before collision 2kg 1 m/s 1 kg

After collision (1/3) m/s

2/3 m/s

CASE 3

Mass of moving object less than the mass of the stationary object (m1 < m2)

(m1 - m2) v’ = (m1 + m2) v1 m1 Moves. But, in the opposite direction.

2m1 v’= (m1 + m2) v1

m2 Moves. V2 has the same direction as V1

Example :

Again, m2 is stationary on the track while the other m1 rolls at 1 m/s towards the right. This time, the mass of m2 is 2 kg while the mass of m1 is 1 kg. What happens when the balls collide elastically?

Since the masses are not equal, we solve again for v1’ and v2’ So

(m1 - m2) 1 kg – 2 kg v’ = (m1 + m2) v1 = 2 kg + 1kg

= 1 m/s = -(1/3) m/s NEGATIVE OPPOSITE DIRECTION 2m1 2 kg v2’= (m1 + m2) v1 = 3 kg

= 1 m/s = (2/3) m/s

Before Collision

m1 1m/s m2 1kg 2kg

After Collision m1 m2 (1/3) m/s 2/3 m/s

TOTAL INELASTIC COLLISION IN ONE DIMENSION

If in an elastic collision the masses maintain their mass state after collision, a total inelastic collision produces a combined mass after collision

Before Collision

1 m/s

After Collision Vf

Formula:m1v1 + m2v2 = (m1 + m2) vf

m1m2

m2m1 +

If m2 is stationary, m2v2 = 0 FORMULA: m1Vf = v1 (m1 + m2)

Example:

Colliding with a much heavier object may produce an inelastic collision. A head-on collision between a car and a train may drag the car along the path of the train. Assume that 1000-kh car stalled on the railroad tracks is smashed by a 50,000-kg train travelling at 8.33 m/s (about 30 km/hr). What is the final velocity after a total inelastic collision?

m1Vf = v1 (m1 + m2)

= 8.33 m/s 50,000 kg 51,000 kg = 8.17 m/s

COLLISIONS IN TWO DIMENSIONS

When we move to collisions in two dimensions, we can separately balance momentum in the x-direction and in the y-direction

Before After

p1’ p2’

a1 a2

ß1 ß2

p1 p2

Using the conservation of energy:

Along the x-axis:

∑px before collision = ∑px after collision (Eqn. 7.16)

p1 cos a1 + p2 cos a2 = p1’ cos ß1 + p2’ cos ß2

Along the y-axis:

∑py before collision = ∑py after collision (Eqn. 7.17)

p1 sin a1 + p2 sin a2 = p1’ sin ß1 + p2’ sin ß2

Example

A pool ball weighing 2 kg is traveling at 30◦ at 0.8 m/s hits another ball moving at 0.5 m/s at 180◦. If the second ball leaves the collision at 0◦ and the first moves away at 150◦, find the final velocity vectors of the balls.

Before After

v’ 1

0.5 m/s ß1

0.8 m/s v’ 2

30◦ ß2 = 0

AnswerGiven

mass of the pool balls m1, m2 2kg

Before Collisionvelocity of the first pool ball v1 0.8m/s

angle of the first pool ball a1 30◦

Velocity of second pool ball v2 0.5 m/s

Angle of second pool ball a2 180◦

After collisionAngle of first pool ball ß1 150◦

Angle of second pool ball ß2 0◦

Find: Final velocities of the balls ( v1’ and v2’)

Solution

Along the x-axis:

∑px before collision = ∑px after collision p1 cos a1 + p2 cos a2 = p1’ cos ß1 + p2’ cos ß2

m(v1 cos a1 + v2 cos a2) + m(v1’ cos ß1 = v2’ cos ß2)

v1 cos a1 + v2 cos a2 = v1’ cos ß1 + v2 cos ß2

0.8 m/s (cos 30◦) + 0.5 m/s (cos 180◦) = v1’ cos 150◦ + v2’ cos 0◦

0.8m/s (0.866) + 0.5 m/s (-1) = v1’ (-0.866) + v2’ (1)

0.70 m/s – 0.5m/s = -0.866 v1’ + v2’

Expressing v2’ in terms of v1’

v2’ = 0.866 v1’ + 0.20m/s

Along the y-axis:

∑py before collision = ∑py after collision

p1 sin a1 + p2 sin a2 = p1’ sin ß1 + p2’ sin ß2

m( v1 sin a1 + v2 sin a2) = m(v1’ sin ß1 + v2’ sin ß2)

v1 sin a1 v2 sin a2 = v1’ sin ß1 + v2’ sin ß2

0.8 m/s (sin 30◦) + 0.5m/s (sin 180◦) = v1’ sin 150◦ + v2 sin 0◦

0.8m/s (0.5) + 0.5m/s (0) = v1’ (0.5) + v2’ (0)

0.4m/s = 0.5 v1’

0.4m/s / 0.5 = v1’

v1’ = 0.m/s < 150◦ Solving for v2’

v2’ = 0.866 v1’ + 0.20 m/s

= 0.866 (0.8m/s) + 0.20 m/s

= 0.9 m/s <0◦