2.7-2.9
Your guide: Sir HamzaBelieves every formula has a story, and units never lie.
Two billiard balls meet. One stops. The other rolls away. Why?
Momentum gives the answer.
Momentum is mass times velocity: p = mv (Eq. 2.25). It is a vector.
The unit is kg m s−1, which is also N s.
From F = ma we get F × t = mv_f − mvᵢ (Eq. 2.26). Force times time equals change in momentum.
This form is more general. It works when mass changes, as in a rocket.
Force times time is called impulse: F × t = mv_f − mvᵢ (Eq. 2.27).
Book Example 2.7: a 1500 kg car slows from 20 to 15 m s−1 in 3.0 s. The force is −2500 N. The minus sign means it is a retarding force.
An isolated system has no outside force on it.
In a collision the two forces are equal and opposite, so F + F′ = 0.
So m1v1 + m2v2 = m1v1′ + m2v2′ (Eq. 2.28).
The book's law: the total linear momentum of an isolated system remains constant.
Momentum is a vector. So give opposite directions opposite signs.
Momentum is conserved in every collision.
In an elastic collision, kinetic energy is also conserved. A hard ball on a marble floor is close to this.
In an inelastic collision, some kinetic energy is lost. It goes to heat, sound or change of shape. A tennis ball dropped on the floor is an example.
Using both laws, the book gets v1 − v2 = −(v1′ − v2′) (Eq. 2.31). The speed of approach equals the speed of separation.
v1′ = (m1 − m2) ÷ (m1 + m2) · v1 + 2m2 ÷ (m1 + m2) · v2 (Eq. 2.32).
v2′ = 2m1 ÷ (m1 + m2) · v1 + (m2 − m1) ÷ (m1 + m2) · v2 (Eq. 2.33).
Four special cases from the book:
(i) Equal masses: the balls swap velocities. v1′ = v2 and v2′ = v1.
(ii) Equal masses and v2 = 0: the first stops. The second moves off at v1. A billiard shot is like this.
(iii) A light body hits a heavy body at rest: v1′ = −v1 and v2′ ≈ 0. The light body bounces back. A squash ball does this.
(iv) A heavy body hits a light body at rest: v1′ ≈ v1 and v2′ ≈ 2v1.
Here the two bodies stick and move together with a common velocity v_f.
m1v1 + m2v2 = (m1 + m2)v_f. So v_f = m1 ÷ (m1 + m2) · v1 + m2 ÷ (m1 + m2) · v2 (Eq. 2.34).
Step 1 / 7
Two billiard balls meet. One stops. The other rolls away. Why?
Momentum gives the answer.
Momentum is mass times velocity: p = mv (Eq. 2.25). It is a vector.
The unit is kg m s−1, which is also N s.
From F = ma we get F × t = mv_f − mvᵢ (Eq. 2.26). Force times time equals change in momentum.
This form is more general. It works when mass changes, as in a rocket.
Force times time is called impulse: F × t = mv_f − mvᵢ (Eq. 2.27).
Book Example 2.7: a 1500 kg car slows from 20 to 15 m s−1 in 3.0 s. The force is −2500 N. The minus sign means it is a retarding force.
An isolated system has no outside force on it.
In a collision the two forces are equal and opposite, so F + F′ = 0.
So m1v1 + m2v2 = m1v1′ + m2v2′ (Eq. 2.28).
The book's law: the total linear momentum of an isolated system remains constant.
Momentum is a vector. So give opposite directions opposite signs.
Momentum is conserved in every collision.
In an elastic collision, kinetic energy is also conserved. A hard ball on a marble floor is close to this.
In an inelastic collision, some kinetic energy is lost. It goes to heat, sound or change of shape. A tennis ball dropped on the floor is an example.
Using both laws, the book gets v1 − v2 = −(v1′ − v2′) (Eq. 2.31). The speed of approach equals the speed of separation.
v1′ = (m1 − m2) ÷ (m1 + m2) · v1 + 2m2 ÷ (m1 + m2) · v2 (Eq. 2.32).
v2′ = 2m1 ÷ (m1 + m2) · v1 + (m2 − m1) ÷ (m1 + m2) · v2 (Eq. 2.33).
Four special cases from the book:
(i) Equal masses: the balls swap velocities. v1′ = v2 and v2′ = v1.
(ii) Equal masses and v2 = 0: the first stops. The second moves off at v1. A billiard shot is like this.
(iii) A light body hits a heavy body at rest: v1′ = −v1 and v2′ ≈ 0. The light body bounces back. A squash ball does this.
(iv) A heavy body hits a light body at rest: v1′ ≈ v1 and v2′ ≈ 2v1.
Here the two bodies stick and move together with a common velocity v_f.
m1v1 + m2v2 = (m1 + m2)v_f. So v_f = m1 ÷ (m1 + m2) · v1 + m2 ÷ (m1 + m2) · v2 (Eq. 2.34).
Q1. A 1500 kg car slows from 20 to 15 m s−1 in 3.0 s. Find the average force.
F × 3.0 = 1500 × 15 − 1500 × 20 = −7500. So F = −2500 N.
✗ “Momentum is conserved only in elastic collisions.”
✓ Momentum is conserved in all collisions. Only kinetic energy depends on the type.
✗ “Ignore signs of velocity.”
✓ Momentum is a vector. Opposite directions need opposite signs.
1. The unit of momentum is:
(b) p = mv gives kg × m s−1.
2. In an elastic collision, which is conserved?
(c) Momentum is always conserved. Elastic also keeps kinetic energy.
3. Equal masses collide elastically. The second is at rest. After the hit:
(a) Book case (ii): v1′ = 0, v2′ = v1.
4. A 2 kg body at 3 m s−1 sticks to a 1 kg body at rest. The common velocity v_f is:
(a) 2 × 3 = (2 + 1) × v_f, so v_f = 2.
Free right now. No ads. Tell us what to fix.