Module 05: Linear Momentum Flashcards
Lesson 5.1 Momentum and Its Conservation
What is Linear Momentum?
Linear Momentum of an object is defined as the product of its mass and its velocity (p)
p = mv
- Is a vector
- The direction of momentum is the direction of velocity
- Unit: kg*m/s
- More momentum - the harder it to stop something
Lesson 5.1 Momentum and Its Conservation
How do you calculate the total momentum of two objects before and after collision?
When we desire to calculate the total momentum of two objects before the collision, their total momentum after collision is zero.
m1u1 + m2u2 = 0
When two objects move in the exact direction, the total momentum before the collision is,
- Pbefore collision = m1u1 + m2u2
When two objects move in opposite directions, the total momentum before the collision is,
- Pbefore collision = m1u1 + -(m2u2)
- Pbefore collision = m1u1- m2u2
Lesson 5.1 Momentum and Its Conservation
How does Newton’s Second Law of Motion apply to collision and momentum?
Force needed to change momentum (increasing/decreasing or changing direction)
🔬 Newton’s Second Law of Motion: The Rate of change of momentum of an object is equal to the net force applied to it
F= Δp/Δt

Lesson 5.1 Momentum and Its Conservation
What is the Law of Conservation of Momentum?
Law of Conservation of Momentum:
The total momentum of an isolated system of objects remains constant
- Momentum is a conserved quantity - when there are no net external forces acting on it
- Hence, as long as there is no external force:
mAvA = mBvB = mAvA’ + mBvB‘
- Must keep time interval small - since nonconserved forces act in the real world
- Conserved as long as PA and PB are measured just before the collision and P’A and P’B just after
- Useful when dealing with simple systems (colliding objects and “explosions”)
Lesson 5.1 Questions
1. (I) What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m/s?

Lesson 5.1 Questions
3. (I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car’s new speed?

Lesson 5.1 Questions
5. (II) Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of 1300 kg/s with a speed of 4.5 × 104 m/s.

Lesson 5.1 Questions
7. (II) A child in a boat throws a 5.30-kg package out horizontally with a speed of 10.0 m/s, Fig. 7– 31. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 24.0 kg and the mass of the boat is 35.0 kg.

Lesson 5.2 Momentum and Its Conservation
What are the forces involved in a collision?
- During a collision of two objects - both objects are deformed because of large forces involved
- Collision: force each exerts on the other usually jumps from zero at the moment of constant to very large force - then returns to zero again
- The magnitude of the force that one object exerts on the other during a collision, as a function of time
- Time is very distinct and very small
- milliseconds for macroscopic collisions
Lesson 5.2 Momentum and Its Conservation
What is the net force of momentum?
Net force (based on Newton’s Second Law): equal to the rate of change of momentum
F = Δp / Δt
- where p = mv
Lesson 5.2 Momentum and Its Conservation
What is impulse?
Applies to each of the two objects in a collision
Impulse: Total change in momentum
- Used when t = small
- Not constant force
- Approximate force using the average force
- Units for impulse: Newton x second (N*s)
FΔt = Δp
Lesson 5.2 Momentum and Its Conservation
15. (I) A 0.145-kg baseball pitched at 31.0 m/s is hit on a horizontal line drive straight back at the pitcher at 46.0 m/s. If the contact time between bat and ball is 5.00 × 10–3 s, calculate the force (assumed to be constant) between the ball and bat.

Lesson 5.2 Momentum and Its Conservation
17. (II) A 12-kg hammer strikes a nail at a velocity of 7.5 m/s and comes to rest in a time interval of 8.0 ms. (a) What is the impulse given to the nail? (b) What is the average force acting on the nail?

Lesson 5.2 Momentum and Its Conservation
19. (II) A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule. (a) What is the change in speed of the space capsule? (b) If the push lasts 0.600 s, what is the average force exerted by each on the other? As the reference frame, use the position of the capsule before the push. (c) What is the kinetic energy of each after the push?

Lesson 5.2 Momentum and Its Conservation
21. (II) A 95-kg fullback is running at 3.0 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west. Calculate (a) the original momentum of the fullback, (b) the impulse exerted on the fullback, (c) the impulse exerted on the tackler, and (d) the average force exerted on the tackler.

Lesson 5.3 Elastic Collisions
When is total kinetic energy conserved?
If two objects are very hard and no heat or energy is produced in the collision = Total Kinetic Energy before and after collision stays constant
Lesson 5.3 Elastic Collisions
What happens to energy the moment of collision?
The moment during contact = energy is stored momentarily in the form of Elastic Potential Energy.
Lesson 5.3 Elastic Collisions
What is an elastic collision?
Elastic Collision is where the total kinetic energy just before and after the collision stays constant
- At the atomic level, the collisions of atoms and molecules are often elastic
- When kinetic energy is not conserved, the total energy is always conserved

Lesson 5.3 Elastic Collisions
What is an inelastic collision?
Inelastic Collision is when the total kinetic energy is not conserved
- Kinetic energy is lost is changing into other forms of energy

Lesson 5.3 Elastic Collisions
Apply the Conservation Laws for Momentum and Kinetic Energy to elastic collisions:
- For v>0 the object is moving to the right (increasing x)
- For v<0 the object is moving to the left (decreasing value of x)

Lesson 5.3 Elastic Collisions
Apply the Conservation Laws for Momentum and Kinetic Energy to elastic collisions’ velocity:
- Know the masses and velocities before
- Determine velocity after collision:
vA + v’A = vB + v’B
Therefore: The relative speed after the collision has the same magnitude of the relative speed before the collision (regardless of mass)
vA - vB = v’B - v’A
Lesson 5.3 Elastic Collisions
25. (II) A ball of mass 0.440 kg moving east (+x direction) with a speed of 3.80 m/s collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

Lesson 5.3 Elastic Collisions
27. (II) A 0.060-kg tennis ball, moving with a speed of 5.50 m/s, has a head-on collision with a 0.090- kg ball initially moving in the same direction at a speed of 3.00 m/s. Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision.

Lesson 5.3 Elastic Collisions
29. (II) A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball. (a) What is the mass of the second ball? (b) What fraction of the original kinetic energy (KE/KE) gets transferred to the second ball?



































