Abelardo M. Zerda IIIMichael O. SuarezJm Dawn C. RivasLeslie Kate Diane Berte
• Momentum and Impulse• Conservation of Momentum• Collisions in One Dimensions• Collisions in Two Dimensions
• Measure of one’s motion, equivalent to the product of one’s mass and velocity.
• Momentum involves motion and mass.• Expressing the definition mathematically, p = mvwhere:p = momentum in kg-m/s unitsm = mass in kg unitsv = velocity of the object in m/s units
• Calculate the momentum of a 100-kg missile traveling at 100 m/s eastward.
• A 100 kg car strikes a tree at 50 km/h. Find its momentum.
• Impulse – force (F) applied by one object to another object within a given time interval Δt
• The formula to get impulse is: F Δt = Δp
where: F Δt = impulse in N-s (newton-second) units Δp = change in momentum in kg-m/s units
• A person in a sled with a total mass of 125 kg, slides down a grassy hill and reach a speed of 8 m/s at the bottom. If a pile of grass can exert a constant force of 250 newtons, how fast will the sled stop?
• A 100 kg car strikes a tree at 20 km/h and comes to a stop in 0.5 seconds. Find its initial momentum and the force on the car while it is being stopped.
• In a collision, energy is not always conserved but all collisions have to conserve momentum if there is no net applied force.
∑ p before collision = ∑ p after collision
There are three types of collisions:
1.Elastic collisions – conserve kinetic energy2.Inelastic collisions – do not conserve kinetic energy3.Completely inelastic collision
• Let us assume that for the two balls, one is running on a straight track toward the second one, which is stationary.
Assume the following items are given:
m1 – mass of object 1m2 – mass of object 2v1 – velocity of object 1 before collisionv2 – velocity of object 2 before collisionv1’ – velocity of object 1 after collisionv2’ – velocity of object 2 after collision
• Solving for the velocity of mass 1 after collision v1’:
v1’ = [(m1 – m2) / (m1 + m2)] v1
• Solving for the velocity of mass 2 after collision v2’:
v2’ = [2m1 / (m1 + m2)] v1
v1’ = [(m1 – m2) / (m1 + m2)] v1
= [(m-m) / (m+m)] v1
= 0After colliding with m2, m1 stops.
v2’ = [2m1 / (m1 + m2)] v1
= (2m / 2m) v1
= v1
Total energy transfer occurred between m1 and m2.
• Since m1 > m2, the term (m1 – m2) is positive.
v1’ = [(m1 – m2) / (m1 + m2)] v1
m1 still moves and v1’ has the same direction as v1.
• Since all the terms are nonzero then,v2’ = [2m1 / (m1 + m2)] v1
m2 also moves and v2’ has the same direction as v1
• Since m1 < m2, the term (m1 – m2) is negative.
v1’ = [(m1 – m2) / (m1 + m2)] v1
m1 still moves but in the opposite direction.
v2’ = [2m1 / (m1 + m2)] v1
Since all the terms are nonzero, then m2 also moves and v2 has the same direction as v1.
Total Inelastic Collisions in One Dimension• Inelastic collisions produce a combined
mass after collision.• The equation for this type of collision is:
m1 v1 + m2 v2 = (m1 + m2) vf
where:
m1 + m2 is the combined mass of the object
Vf is the final velocity value for the combined mass
If m2 v2 = 0, vf = v1 m1/ (m1 + m2)
• Balanced momentum in the x-direction and y-direction
• Angles are arbitrary
• Along the x-axis:
px before collision = px after collision
p1 cos 1 + p2 cos 2 = p1 cos 1 + p2 cos 2
• Along the y-axis:
py before collision = py after collision
p1 sin 1 + p2 sin 2 = p1 sin 1 + p2 sin 2
• 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.
