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Lecture 23:

Collisions and conservation of momentum II

Objectives

1. Solve problems involving systems in 1D and 2D

where linear momentum is conserved.

Completely Inelastic Collisions

2

(𝒎𝑨 + 𝒎𝑩)𝒗𝟐 = 𝒎𝑨𝒗𝑨𝟏 + 𝒎𝑩𝒗𝑩𝟏

Inelastic collisions

Non-conservative internal forces present

Dissipates energy

Completely inelastic collisions

Momentum is conserved

Final velocity common for the entire system

Objects stick to each other!

Note: Same equation yesterday I just replaced (1) and (2) with A and B, and initial (i) and final (f) with 1 and 2. You can use whatever notation you are comfortable with as long as the substitution are consistent. 

Steps in solving collision problems:

1. Dissect the problem

• illustrate before and after collisions
• list all the given (be careful with signs!)

2. If 2-D or 3-D, decompose the velocity/momenta in x-, y-

and z-components.

3. Determine if the collision if elastic or inelastic.

4. Use proper sets of equations and do the math:

𝑚 1 𝑣1𝑓^2 +

𝑚 2 𝑣2𝑓^2 =

𝑚 1 𝑣1𝑖^2 +

𝑚 2 𝑣2𝑖^2

Elastic

collisions

Completely

Inelastic

collisions

Quick Examples: (no calculators)

Two identical masses A and B undergoes perfect elastic

collision. What will be the velocity of B after collision for

the following situation:

I. Mass B is initially at rest and A is moving 1.0m/s East;

after collision ball A stopped moving?

Sample Problem: Young and Freedman, 8.42: A 0.150 kg glider

is moving to the right on a frictionless horizontal air track with a

speed of 0.800 m/s. It has a head on collision with a 0.300 kg

glider that is moving to the left with a speed of 2.20 m/s. Find

(a) final velocity (magnitude and direction) of each glider if the

collision is elastic and (b) the change in KE.

Given: mA = 0.150kg mB = 0.300kg vA1x = 0.80m/s vB1x = - 2.20m/s Required: vA2x and vB2x

Using conservation of total momentum in x-component to get vB2x:

2 unknowns (vA2x, vB2x)

− 𝑣𝐵2𝑥 − 𝑣𝐴2𝑥 = 𝑣𝐵1𝑥 − 𝑣𝐴1𝑥 rel. velocity

Using the relative velocities for elastic collisions:

Substituting vB2x to va2x :

Given: mA = 0.150kg mB = 0.300kg vA1x = 0.800m/s vB1x = - 2.20m/s Required: vA2x and vB2x

𝑣𝐴2𝑥 + 𝑚𝐴 𝑚𝑣𝐴2𝑥 𝐵

= 2𝑣𝐵1𝑥 − 𝑣𝐴1𝑥 + 𝑚𝐴 𝑚𝑣𝐴1𝑥 𝐵

→ 𝑣𝐴2𝑥 1 + 𝑚𝑚𝐴 𝐵

= 2𝑣𝐵1𝑥 − 𝑣𝐴1𝑥 + 𝑚𝐴 𝑚𝑣𝐴1𝑥 𝐵

𝑣𝐴2𝑥 =

2𝑣𝐵1𝑥 − 𝑣𝐴1𝑥 + 𝑚𝐴 𝑚𝑣𝐵𝐴1𝑥 1 + 𝑚 𝑚𝐴𝐵

=

2 − 2.2𝑚𝑠 − 0.80𝑚𝑠 + (0.15𝑘𝑔)(0.80𝑚/𝑠)0.30𝑘𝑔

1 + 0.15𝑘𝑔0.30𝑘𝑔

Given:

mA = 20kg

mB = 12kg

vB1 =

vA1 = 2.0m/s

vA2 = 1.0m/s

Required: vB

Sample Problem:

Two battling robots sliding on a frictionless surface. Robot A,

with mass 20kg, initially moves at 2.0m/s parallel to the x-axis.

It collides with robot B, which has mass 12kg and is initially at

rest. After the collision, robot A is moving at 1.0m/s in a

direction that makes an angle α = 30o^ with its initial direction.

What is the final velocity of robot B?

Final velocity vB2 has x and y components: We must get for vB2x and vB2y to solve for vB

Apply conservation of momentum in x component to get vB2x

Apply conservation of momentum in y component to get vB2y

𝑣𝐴2𝑥 = 𝑣𝐴2𝑐𝑜𝑠𝛼 𝑣𝐴2𝑦 = 𝑣𝐴2𝑠𝑖𝑛𝛼

After the collision, robot B moves in the x-positive direction and negative y direction: the magnitude of vB2 is:

𝑣𝐵2 = (𝑣𝐵2𝑥)^2 + (𝑣𝐵2𝑦)^2

𝑣𝐵2 = (1.89𝑚/𝑠)^2 +(−0.83𝑚/𝑠)^2 = 𝟐. 𝟏𝒎/𝒔

The angle it makes with the horizontal axis is:

Sample problem: An 0.50-kg object (A) is compressing the

spring (k = 10N/m) by 0.20m. The spring is attached to a wall

as shown. After the object is released, it collides with a 0.10-

kg ball that is moving at 1.0m/s at 45o, North of West. What

is the final velocity the ball after collision if object A moves

0.2m/s East? (Surface friction is negligible)

A

B

State 1: object A compressing the spring State 2: before collision (after the spring released the ball) State 3: after collision Given: mA, mB, k, x, vB2x, vB2y vA3x,

Required: vB

top view

Given: mA = 0.5kg mB = 0.1kg vB2 = 1.0m/s (45o) vA2x = 0.89m/s vA2y =0; vA3y = vA3x = 0.20m/s vB3 =???

𝑣𝐵2𝑥 =

0.5𝑘𝑔 0.89𝑚𝑠 − 0.1𝑘𝑔 1.0𝑚𝑠 𝑐𝑜𝑠45 − (0.5𝑘𝑔)(0.2𝑚𝑠 ) 0.1𝑘𝑔 = 𝟐. 𝟕𝐦/𝐬

𝑣𝐵2𝑦 =

0.5𝑘𝑔 0 + 0.1𝑘𝑔 1.0𝑚𝑠 𝑠𝑖𝑛45 − (0.5𝑘𝑔)(0) 0.1𝑘𝑔

= 𝟎. 𝟕𝟏𝒎/𝒔

For the x-component the conservation of total momentum gives:

For the y-component the conservation of total momentum gives:

After the collision, tha ball B moves at velocity:

𝑣𝐵3 = (𝑣𝐵3𝑥)^2 + (𝑣𝐵3𝑦)^2

𝑣𝐵3 = (2.7𝑚/𝑠)^2 +(0.71𝑚/𝑠)^2 ~𝟐. 𝟖𝒎/𝒔

The angle it makes with the horizontal axis is:

Homework Answers

Homework (1,2)