Chapter 4 Impulse and Momentum

Conservation of Momentum

Momentum can be defined as "mass in motion". All objects have mass, so if an object is moving, then it has momentum. You can consider momentum to be an object’s resistance to stopping. It's a kind of inertia for moving an object.

Considering a collision between two objects, and according to Newton’s third law, the forces acting between the two objects are equal in magnitude but opposite in direction. Since two objects experience the collision within the same period of time, the impulses acting between the two objects are also equal in magnitude but opposite in direction. Therefore, the momentum changed will be equal in magnitude but opposite in direction. The overall momentum will not be changed and the total momentum is conserved.

Law of Conservation of Momentum

Momentum is conserved in a system if no external force acts upon it. The total momentum before the events, such as collisions or explosions, is equal to the total momentum after the events.

1. The key of solving momentum problems is to find the total momentum before and after and to solve them algebraically by setting them equal.

2. Elastic and Inelastic Collisions

An elastic collision is defined as one where there is no loss of kinetic energy in the collision. Both momentum and kinetic energy are conserved during the elastic collisions. An inelastic collision occurs when two colliding objects stick together after the impact. Only momentum is conserved during an inelastic collision.

3. When object A and object B collide elastically:

Where are the relative velocities of object A and B to the velocity of the center of mass.

Velocity of the center of mass is constant under no external force condition.

Two-Dimensional Collision

Since momentum is a vector quantity, you need to set up conservation of momentum in both the horizontal (x−direction) and the vertical (y−direction) directions.

Examples:

1. A 0.01-kg bullet moving horizontally at 200 m/s crashes into a stationary 2.99-kg block on a level and horizontal surface.

a. What is the speed of the bullet/block combination after the perfectly inelastic collision?

Answer: The total momentum before is equal to the total momentum after.

0.01 × 200 + 0 = (0.01 + 2.99) × v

b. How much energy is lost in the collision?

Answer: Energy Lost = Energy Before — Energy After

Energy Lost = =199.33 J

2. A bullet with a mass of 0.008 kg (8.0 grams) is fired horizontally into a stationary 9.0-kg block of wood on a frictionless horizontal surface. The bullet sticks into the block of wood and the block has a speed of 0.4 m/s after the impact. What is the initial velocity of the bullet?

Answer: Use the law of conservation of momentum.

0.008 × v + 0 = (9 + 0.008) × 0.4

v = 450.4 m/s

3. A 4-kg block moving at 20 m/s collides elastically, head-on, with an initially stationary 6-kg block. What are the speeds and directions of the two blocks after the collision?

Answer: Velocity of Center of Mass = = 8 m/s

For 4-kg block:

m/s

For 6-kg block: 0—8 += 0

m/s

4. A bullet with a mass of 0.050 kg and an initial velocity of 200 m/s is shot horizontally at a 2-kg block suspended as shown below. The bullet becomes imbedded in the block.

a. Calculate the velocity of the bullet-block system just after the impact.

Answer: Use the law of conservation of momentum.

0.05 × 200 + 0 = (2 + 0.05) × v’

v’ = 4.88 m/s

b. Calculate the kinetic energy of the bullet-block system just after the impact.

Answer:

c. Calculate the maximum height to which the bullet-block system swings.

Answer: Mechanical energy is conserved after the impact.

24.39 = 2.05 × 9.8 × h

h = 1.21 m

5. A 1-kg air hockey puck traveling at 2 m/s collides with a 2-kg puck. The two pucks move off after the collision as indicated in the diagram below. Determine the final velocity of each puck.

Answer: Use conservation of momentum for both x and y directions.

For x—direction: 2 × 1 + 0 =

For y—direction: 0 =

and