Momentum – Part 2

Conservation of Momentum

Conservation of Momentum

The law of conservation of momentum tells us that as long as colliding objects are not influenced by outside forces like friction, the total amount of momentum in the system before and after the collision is the same.

Example #1

What is the momentum of a 0.2-kilogram steel ball that is rolling at a velocity of 3.0 m/s?

momentum = mass × velocity P = m × v

P

Example #2You and a friend stand facing each other on ice skates. Your mass is 50. kilograms and your friend’s mass is 60. kilograms. As the two of you push off each other, you move with a velocity of 4.0 m/s to the right.Looking for:

your friend’s velocity to the leftGiven:

Your mass of 50 kgYour friend’s mass of 60 kgYour velocity of 4.0 m/s to the right

Relationship:m1v3 = - (m2v4)

m1v3 = - (m2v4)(50 kg)(4m/s) = - (60 kg) v4

(200 kgm/s) = - (60 kg) v4

Solve for v4:

𝑉 4=200𝑘𝑔 •𝑚/ 𝑠−60𝑘𝑔

- 3.3 m/s (this is left)

Example #3 Finding Force

• Force is measured in Newtons (N), which has the real units of kg•m/s2.

• We know that F = ma, and a = Δv/t• This tells us that F = m v/t• F t = mv, and mv = P, so P is also = F t• This means that a change in momentum

divided by time, give an average force• F = ΔP/t

Example of Force

An arrow is shot into a straw target. The arrow has a mass of 0.2kg, and is moving at a velocity of 150 m/s. It is stopped over a time of 0.01 second. What force did the target apply to the arrow?P = m × vP = (0.2 kg)(150 m/s)P= 30 kg•m/s 𝐹=

30𝑘𝑔 •𝑚 /𝑠0.01𝑠

𝐹=Δ 𝑃𝑡

= 3000 N

Example of Distance

The same arrow is shot 0.75 m into the straw target. The arrow has a mass of 0.2kg, and is moving at a velocity of 150 m/s. What force did the target apply to the arrow? (similar to question #7b)