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.
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 - ?
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)
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
If in an elastic collision the masses maintain their mass state after collision, a total inelastic collision produces a combined mass after collision
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?
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◦