College Physics I: BIIG problem-solving method
Linear Momentum
- Linear momentum is defined as the product of a system’s mass multiplied by its velocity
p = m v
The SI unit for momentum is kg m/s.
- Momentum is directly proportional to the object’s mass and also its velocity.
- Thus the greater an object’s mass or the greater its velocity, the greater its momentum.
- Momentum p is a vector having the same direction as the velocity v.
Momentum and Newton’s Second Law
- We know that the Newton’s Second Law is given by the force
Fnet = m a
Using the definition of acceleration
a = Δv / Δt
Therefore,
Fnet = m ( Δv / Δt )
- Newton’s Second Law of Motion in Terms of Momentum: The net external force equals the change in momentum of a system divided by the time over which it changes.
Fnet = Δp / Δt
where, Fnet is the net external force, Δp is the change in momentum, and Δt is the change in time.
- Problem (E8.2): During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by the racquet, assuming that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)? ( 660 N )
Impulse
- The net external force is
Fnet = Δp / Δt
- Impulse is a very large force acting for a short time.
- Change in momentum equals the average net external force multiplied by the time this force acts.
Δp = Fnet Δt
Impulse is the same as the change in momentum.
Isolated System
- An isolated system is defined to be one for which the net external force is zero
Fnet = 0
Conservation of Momentum
- According to the impulse-momentum theorem, the changes in the momenta of the two objectsalong one directionare
∆p1 = J1 and ∆p2 = J2
From Newton’s third law, the action/reaction pairs are opposite, so
J1 = - J2
Therefore,
∆p1 = - ∆p2
Rearranging, we get
∆p1 + ∆p2 = ∆(p1 + p2) = 0
This means,
p1 + p2 = constant
Therefore, the sum of the momenta after the collision equals the sum of the momenta before the collision.
(p1 + p2)f = (p1 + p2)i = constant
In terms of total momentum
Pf = Pi = constant (Does not change during collision!)
We conclude, the total momentum is conserved.
- Using Newton’s Second law in terms of momentum,
Fnet = Δp / Δt
For an isolated system,
Fnet = 0
Therefore
Δp = 0 and p = constant
- Conservation of momentum principle: The total momentum is conserved for any isolated system.
- Conservation of momentum is violated only when the net external force is not zero.
- The conservation of momentum principle not only applies to the macroscopic objects, it is also essential to our explorations of atomic and subatomic particles.
Elastic Collisions
- Internal kinetic energy is the sum of the kinetic energies of the objects in the system.
- Elastic Collision: An elastic collision is one that conserves internal kinetic energy.
- Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei.
- Momentum and internal kinetic energy are conserved.
- Problem
(E8.4): For an elastic collision,
given that
m1 = 0.500 kg, m2 = 3.50 kg, v1 = 4.00 m/s, and v2 = 0
If the velocity of the first object after the collision is 3.00 m/s in the opposite direction, calculate the velocity of the second object after the collision. ( 1.00 m/s )
Inelastic Collisions
- An inelastic collision is one in which the internal kinetic energy changes (it is not conserved).
- Problem (E8.5): A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. If goalie’s recoil velocity is 0.0748 m/s, how much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. ( - 91.7 J )
Point Mass
- One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision.
- Point mass is a structureless particle, that cannot rotate or spin.
Conservation of Momentum
- The approach taken in discussing two-dimension is to choose a convenient coordinate system and
resolve the motion into components along perpendicular axes.
- Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously.
- Conservation of Momentum along the x -axis
m1 v1 = m1 v′1 cos θ1 + m2 v′2 cos θ2
Conservation of Momentum along the y -axis
m1 v1 = m1 v′1 sin θ1 + m2 v′2 sin θ2
Elastic Collisions of Two Objects with Equal Mass
- Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic.
- This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions.
- There are three ways possibilities:
head-on collision; incoming ball stops
no collision; incoming ball continues unaffected
angle of separation is 90º after the collision.
Acceleration of a Rocket
- Acceleration of a rocket is
a = [ ( ve / m ) ( Δmb / Δt ) ] − g
where, a is the acceleration of the rocket, ve is the escape velocity, m is the mass of the rocket,
Δmb is the mass of the ejected gas, and Δt is the time in which the gas is ejected.
- Factors Affecting a Rocket’s Acceleration: exhaust velocity of the gases relative to the rocket,
the burning of the rocket fuel, the rocket’s mass.
- The rocket mass m decreases dramatically during flight because most of the rocket is fuel to begin with, so that acceleration increases continuously, reaching a maximum just before the fuel is exhausted.
Rocket Propulsion
- Newton’s third law of motion: Matter is forcefully ejected from a system, producing an equal and opposite reaction on what remains.
- Examples: The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses.
- Problem (E8.8): A Saturn V’s mass at liftoff was 2.80 × 106 kg , its fuel-burn rate was 1.40×104 kg/s, and the exhaust velocity was 2.40 × 103 m/s. Calculate its initial acceleration. ( 2.20 m/s2 )