a summary textbook for physics of classical

A SUMMARY TEXTBOOK FOR PHYSICS OF CLASSICAL MECHANICS 101

Written by Waleed Sh. Abu Khader Reviewed by Saleem Abdo

University of Jordan School of Engineering First Edition 2015

Contents Chapter 12 : Collisions 12.1 Collisions in One Dimension

2

12.2 Collisions in Two Dimensions

6

Problems

8

‫‪Chapter Twelve : Collisions‬‬

‫‪12.1 Collisions in One Dimension‬‬

‫‪Types of collisions :‬‬

‫ال مرن تماما‬

‫مرن‬

‫ال مرن‬

‫يتالصق الجسمان‬ ‫المتصادمان و‬ ‫يتابعان سيرهما‬ ‫بسرعة واحد كجسم‬ ‫واحد‬ ‫يرتد الجسمان بعد‬ ‫تصادمهما و‬ ‫يتحركان بشكل‬ ‫منفصل‬ ‫يتغير شكل الجسمين‬ ‫بعد تصادمهما بحيث‬ ‫تنقص طاقة الحركة‬ ‫بعد التصادم إال‬ ‫أنهما يتابعان‬ ‫الحركة منفصلين‬

‫‪-‬‬

‫حفظ كمية الحركة‬ ‫( الزخم ) فقط‬

‫حفظ كمية الحركة‬ ‫( الزخم ) و‬ ‫الطاقة الحركية‬

‫حفظ كمية الحركة‬ ‫( الزخم ) فقط‬

‫‪ Perfectly Inelastic Collisions :‬‬ ‫𝝑( 𝟐𝒎 ‪⃗⃗ 𝟏 )𝒊 +‬‬ ‫𝝑( ) 𝟐𝒎 ‪⃗ 𝟐 )𝒊 = (𝒎𝟏 +‬‬ ‫𝒇)𝟐‪⃗ 𝟏,‬‬ ‫𝝑( 𝟏𝒎‬

‫‪Example 12.1.1 : A car of mass ( 𝟗𝟕𝟓 𝒌𝒈 ) moving with ( 𝟐𝟐 𝒎/𝒔 ) collides with a truck of‬‬ ‫‪mass ( 𝟏𝟖𝟓𝟎 𝒌𝒈 ) that rests on a signal and sticks with it and continues moving as a one‬‬ ‫‪object , Calculate the speed of that object .‬‬

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‫وليد شاهر أبوخضر‬

⃗⃗ 𝟏 )𝒊 + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒊 = (𝒎𝟏 + 𝒎𝟐 ) (𝝑 ⃗ 𝟏,𝟐)𝒇 𝒎𝟏 (𝝑 ⃗ 𝟏,𝟐 )𝒇 𝟎 + 𝟗𝟕𝟓 × 𝟐𝟐 = (𝟗𝟕𝟓 + 𝟏𝟖𝟓𝟎)(𝝑 ⃗⃗ 𝟏,𝟐 )𝒇 = 𝟕. 𝟓𝟗 𝒎/𝒔 (𝝑 → 𝒊𝒏 𝒕𝒉𝒆 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒐𝒗𝒊𝒏𝒈 𝒄𝒂𝒓

Example 12.1.2 : A red car of mass ( 𝟐𝟏𝟎𝟎 𝒌𝒈 ) moving with ( 𝟏𝟏 𝒋̂ 𝒎/𝒔 ) , collides with a yellow car of mass ( 𝟑𝟏𝟎𝟎 𝒌𝒈 ) , if both cars moved with ( − 𝟓 𝒋̂ 𝒎/𝒔 ) as a one object after collision , find the velocity of the yellow car before the collision .

⃗⃗ 𝟏 )𝒊 + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒊 = (𝒎𝟏 + 𝒎𝟐 ) (𝝑 ⃗ 𝟏,𝟐)𝒇 𝒎𝟏 (𝝑 ⃗ 𝟐 ) = (𝟑𝟏𝟎𝟎 + 𝟐𝟏𝟎𝟎) × − 𝟓 𝒋̂ 𝟐𝟏𝟎𝟎 × 𝟏𝟏𝒋̂ + 𝟑𝟏𝟎𝟎(𝝑 𝒊 ⃗⃗ 𝟐 ) = −𝟏𝟓. 𝟖 𝒋̂ 𝒎/𝒔 (𝝑 𝒊

-

Inelastic Collisions : . ‫ نقص في الطاقة الحركية‬ . ‫ أشكال األجسام تتغير و تفقد جزء من طاقها‬ . ‫ مع ذلك يتم الحفاظ على كمية الحركة‬

Example 12.1.3 : Two balls of clay collides a perfectly inelastic collision ( 𝒎𝟏 = 𝟎. 𝟓 𝒌𝒈 , 𝝑𝟏𝒊 = 𝟒 𝒊̂ 𝒎/𝒔 ) and ( 𝒎𝟐 = 𝟎. 𝟐𝟓 𝒌𝒈 , 𝝑𝟐𝒊 = −𝟑 𝒊̂ 𝒎/𝒔 ) , Determine : I. II.

The final velocity of the combined mass of clay after collision . The change in kinetic energy due collision .

I. ⃗⃗ 𝟏 )𝒊 + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒊 = (𝒎𝟏 + 𝒎𝟐 ) (𝝑 ⃗ 𝟏,𝟐)𝒇 𝒎𝟏 (𝝑 ‫وليد شاهر أبوخضر‬

3|Page

‫𝒇) 𝟐‪⃗⃗ 𝟏,‬‬ ‫𝝑()𝟓 ‪𝟎. 𝟓 × 𝟒 𝒊̂ + 𝟎. 𝟐𝟓 × −𝟑 𝒊̂ = (𝟎. 𝟐𝟓 + 𝟎.‬‬ ‫𝒔‪⃗ 𝟏,𝟐)𝒇 = 𝟏. 𝟔𝟕 𝒊̂ 𝒎/‬‬ ‫𝝑(‬

‫‪II.‬‬ ‫𝟐𝒊𝑬𝑲 ‪𝑲𝑬𝒊 = 𝑲𝑬𝒊𝟏 +‬‬ ‫𝟏‬ ‫) 𝒊𝟐𝟐𝝑 𝟏𝒎 ‪= (𝒎𝟏 𝝑𝟐𝟏𝒊 +‬‬ ‫𝟐‬ ‫𝑱 𝟐𝟏 ‪= 𝟓.‬‬ ‫𝟐𝒇𝑬𝑲 ‪𝑲𝑬𝒇 = 𝑲𝑬𝒇𝟏 +‬‬ ‫𝟏‬ ‫)𝒇𝟐𝟐𝝑 𝟏𝒎 ‪= (𝒎𝟏 𝝑𝟐𝟏𝒇 +‬‬ ‫𝟐‬ ‫𝑱 𝟓𝟎 ‪= 𝟏.‬‬ ‫𝒊𝑬𝑲 ‪∆𝑲𝑬 = 𝑲𝑬𝒇 −‬‬ ‫𝑱 𝟕𝟎 ‪= −𝟒.‬‬

‫‪Elastic Collisions :‬‬

‫‪-‬‬

‫‪ ‬القوى غير ثابتة أثناء التصادمات الحقيقية لكنها‬ ‫متساوية عند أي لحظة معينة خالل التصادم ‪،‬‬ ‫ذلك حسب قانون نيوتن الثالث ‪.‬‬ ‫‪ ‬كمية الحركة محفوظة ‪.‬‬ ‫‪ ‬الطاقة الحركية محفوظة ( 𝟎 = 𝑬𝑲∆ ) ‪.‬‬

‫‪‬‬ ‫𝝑( 𝟐𝒎 ‪⃗⃗ 𝟏 )𝒊 +‬‬ ‫𝝑( 𝟏𝒎 = 𝒊) 𝟐 ⃗‬ ‫𝝑( 𝟐𝒎 ‪⃗ 𝟏 )𝒇 +‬‬ ‫𝒇) 𝟐 ⃗‬ ‫𝝑( 𝟏𝒎‬ ‫‪‬‬ ‫𝟏‬ ‫𝟏‬ ‫𝟏‬ ‫𝟏‬ ‫𝟐 𝟐𝒇𝝑 𝟐𝒎 ‪𝒎𝟏 𝝑𝒊𝟏 𝟐 + 𝒎𝟐 𝝑𝒊𝟐 𝟐 = 𝒎𝟏 𝝑𝒇𝟏 𝟐 +‬‬ ‫𝟐‬ ‫𝟐‬ ‫𝟐‬ ‫𝟐‬ ‫‪ Since the second equation is mathematically hard to manage with two‬‬ ‫‪variables we can use the relative velocity equation instead :‬‬ ‫𝝑(‪⃗𝝑𝟏𝒊 − ⃗𝝑𝟐𝒊 = −‬‬ ‫)𝒇𝟐𝝑⃗ ‪⃗⃗ 𝟏𝒇 −‬‬ ‫‪4|Page‬‬

‫وليد شاهر أبوخضر‬

Example 12.1.4 : Two balls of glasses ( 𝒎𝟏 = 𝟎. 𝟎𝟏𝟓 𝒌𝒈 , 𝒎𝟐 = 𝟎. 𝟎𝟑 𝒌𝒈 ) where they move with ( ⃗⃗𝝑𝟏𝒊 = 𝟎. 𝟐𝟐𝟓 𝒊̂ 𝒎/𝒔 , ⃗𝝑𝟐𝒊 = −𝟎. 𝟏𝟖 𝒊̂ 𝒎/𝒔 ) , collides head-on collision , if ball ( 𝟏 ) bounces to the left with speed ( 𝟎. 𝟑𝟏𝟓 𝒎/𝒔 ) . Determine : I. II.

The final velocity of the ball ( 𝟐 ) . Whether the kinetic energy of the system is conserved or not .

I. ⃗⃗ 𝟏 )𝒊 + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒊 = 𝒎𝟏 (𝝑 ⃗ 𝟏 )𝒇 + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒇 𝒎𝟏 (𝝑 ⃗ 𝟐 )𝒇 𝟎. 𝟎𝟏𝟓 × 𝟎. 𝟐𝟐𝟓 𝒊̂ + 𝟎. 𝟎𝟑 × −𝟎. 𝟏𝟖 𝒊̂ = 𝟎. 𝟎𝟏𝟓 × −𝟎. 𝟏𝟑𝟓 𝒊̂ + 𝟎. 𝟎𝟑 × (𝝑 ⃗⃗ 𝟐 )𝒇 = 𝟗 × 𝟏𝟎−𝟐 𝒊̂ 𝒎/𝒔 (𝝑

II. 𝟏 𝟏 𝒎𝟏 𝝑𝒊𝟏 𝟐 + 𝒎𝟐 𝝑𝒊𝟐 𝟐 = 𝟖. 𝟕 × 𝟏𝟎−𝟒 𝑱 𝟐 𝟐 𝟏 𝟏 𝑲𝑬𝒇 = 𝒎𝟏 𝝑𝒇𝟏 𝟐 + 𝒎𝟐 𝝑𝒇𝟐 𝟐 = 𝟖. 𝟕 × 𝟏𝟎−𝟒 𝑱 𝟐 𝟐 ∆𝑲𝑬 = 𝟎 → 𝑪𝒐𝒏𝒔𝒆𝒓𝒗𝒆𝒅 𝑲𝑬𝒊 =

Example 12.1.5 : A block of mass ( 𝒎𝟏 = 𝟏. 𝟔 𝒌𝒈 ) , initially moving to the right with speed ( 𝟒 𝒎/𝒔 ) , collides elastically with a block of mass ( 𝟐. 𝟎𝟏 𝒌𝒈 ) moving opposite to block ( 𝟏 ) with ( 𝟐. 𝟓 𝒎/𝒔 ) , Calculate the velocity of each block after collision . 𝑨𝒏𝒔 ∶ ⃗⃗ 𝟏 )𝒊 + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒊 = 𝒎𝟏 (𝝑 ⃗ 𝟏 )𝒇 + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒇 𝒎𝟏 (𝝑 ⃗ 𝟏 ) + 𝟐. 𝟎𝟏(𝝑 ⃗⃗ 𝟐 )𝒇 𝟏. 𝟔 × 𝟒 + 𝟐. 𝟎𝟏 × −𝟐. 𝟓 = 𝟏. 𝟔(𝝑 𝒇 →

𝟏𝟏 ⃗ 𝟏 ) + 𝟐. 𝟎𝟏(𝝑 ⃗ 𝟐 )𝒇 →→→ (∗) = 𝟏. 𝟔(𝝑 𝒇 𝟖 ⃗ 𝟏𝒊 − 𝝑 ⃗ 𝟐𝒊 = −(𝝑 ⃗⃗ 𝟏𝒇 − 𝝑 ⃗ 𝟐𝒇) 𝝑 ⃗ 𝟏𝒇 − 𝝑 ⃗ 𝟐𝒇) 𝟒 − (−𝟐. 𝟓) = −(𝝑

‫وليد شاهر أبوخضر‬

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→ ⃗𝝑𝟏𝒇 = −𝟔. 𝟓 + ⃗𝝑𝟐𝒇 →→ (∗∗) Substitute ( ** in * ) : →

𝟏𝟏 ⃗ 𝟐 )𝒇 = 𝟏. 𝟔 × (−𝟔. 𝟓 + ⃗𝝑𝟐𝒇) + 𝟐. 𝟎𝟏(𝝑 𝟖 ⃗ 𝟐 )𝒇 = 𝟑. 𝟐𝟔 𝒎/𝒔 (𝝑 ⃗⃗ 𝟏 )𝒇 = −𝟑. 𝟐𝟒 𝒎/𝒔 (𝝑

12.2 Collisions in Two Dimensions 𝒙

𝒚

⃗ 𝟏 )𝒊𝒙 + 𝒎𝟐 (𝝑 ⃗ 𝟐)𝒊𝒙 = 𝒎𝟏 (𝝑 ⃗ 𝟏 )𝒇𝒙 + 𝒎𝟐 (𝝑 ⃗ 𝟐)𝒇𝒙 𝒎𝟏 (𝝑

⃗ 𝟏)𝒊𝒚 + 𝒎𝟐 (𝝑 ⃗ 𝟐)𝒊𝒚 = 𝒎𝟏 (𝝑 ⃗ 𝟏 )𝒇𝒚 + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒇𝒚 𝒎𝟏 (𝝑

⃗ 𝟏)𝒊𝒙 + 𝟎 = 𝒎𝟏 (𝝑 ⃗ 𝟏)𝒇 𝒄𝒐𝒔∅ + 𝒎𝟐 (𝝑 ⃗ 𝟐)𝒇 𝒄𝒐𝒔𝜽 𝒎𝟏 (𝝑

⃗ 𝟏 )𝒊𝒚 = (𝝑 ⃗ 𝟐 )𝒊𝒚 = 𝟎 (𝝑

⃗ )𝒊𝒙 = (𝝑 ⃗ )𝒊 (𝝑

𝟎 = 𝒎𝟏 (𝝑⃗ 𝟏)𝒇 𝒔𝒊𝒏∅ + 𝒎𝟐 (𝝑⃗ 𝟐 )𝒇𝒔𝒊𝒏𝜽 ) 𝝑𝒇𝒚𝟐 ( ‫إتجاه ( 𝟏𝒚𝒇𝝑 ) عكس إتجاه‬

⃗ 𝟏 )𝒊𝒙 = 𝒎𝟏 (𝝑 ⃗ 𝟏)𝒇 𝒄𝒐𝒔∅ + 𝒎𝟐 (𝝑 ⃗ 𝟐 )𝒇 𝒄𝒐𝒔𝜽 𝒎𝟏 (𝝑

𝟎 = 𝒎𝟏 (𝝑𝟏)𝒇 𝒔𝒊𝒏∅ − 𝒎𝟐 (𝝑𝟐 )𝒇𝒔𝒊𝒏𝜽

Remarks : -

If the collision is elastic we could apply the conservation of kinetic energy : 𝟏 𝟏 𝟏 𝒎𝟏 𝝑𝒊𝟏 𝟐 = 𝒎𝟏 𝝑𝒇𝟏 𝟐 + 𝒎𝟐 𝝑𝒇𝟐 𝟐 𝟐 𝟐 𝟐 𝒘𝒉𝒆𝒓𝒆 𝝑𝒊𝟐 = 𝟎

-

Applying the relative velocity equation is incorrect in the [ 2D ] collisions : ⃗𝝑𝟏𝒊 − ⃗𝝑𝟐𝒊 = −(𝝑 ⃗⃗ 𝟏𝒇 − ⃗𝝑𝟐𝒇)

‫وليد شاهر أبوخضر‬

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-

Assuming that :  ( 𝒎𝟏 = 𝒎𝟐 ) ⃗ 𝟐 )𝒊 = 𝟎  (𝝑 

Elastic collision

Then : 

∅ + 𝜽 = 𝟗𝟎°

Example 12.2.1 : An object of mass ( 𝟑 𝒌𝒈 ) is moving with an initial velocity of ( 𝟓 𝒊̂ 𝒎/𝒔 ) , collides with another object of mass ( 𝟑 𝒌𝒈 ) at rest . If the moving object veered his direction after this elastic collision with an angle of ( 𝟓𝟑° ) above the ( +𝒙 − 𝒂𝒙𝒊𝒔 ) . Calculate the final speed of each object . 𝒙

𝒚

⃗⃗ 𝟏 )𝒊𝒙 = 𝒎𝟏 (𝝑 ⃗⃗ 𝟏 )𝒇 𝒄𝒐𝒔∅ + 𝒎𝟐 (𝝑 ⃗⃗ 𝟐 )𝒇 𝒄𝒐𝒔𝜽 𝒎𝟏 (𝝑

𝟎 = 𝒎𝟏 (𝝑𝟏 )𝒇𝒔𝒊𝒏∅ − 𝒎𝟐 (𝝑𝟐 )𝒇𝒔𝒊𝒏𝜽

𝒎𝟏 = 𝒎𝟐 , 𝒆𝒍𝒂𝒔𝒕𝒊𝒄 𝒄𝒐𝒍𝒍𝒊𝒔𝒊𝒐𝒏 → ∅ + 𝜽 = 𝟗𝟎 ⃗⃗ 𝟏 )𝒊𝒙 = (𝝑 ⃗ 𝟏 )𝒇𝒄𝒐𝒔∅ + (𝝑 ⃗ 𝟐 ) 𝒄𝒐𝒔(𝟗𝟎 − ∅) (𝝑 𝒇

𝟎 = (𝝑𝟏 )𝒇 𝒔𝒊𝒏∅ − (𝝑𝟐 )𝒇𝒔𝒊𝒏(𝟗𝟎 − ∅)

⃗⃗ 𝟏 )𝒇𝒄𝒐𝒔𝟓𝟑 + (𝝑 ⃗ 𝟐 ) 𝒄𝒐𝒔(𝟗𝟎 − 𝟓𝟑) 𝟓 = (𝝑 𝒇

𝟎 = (𝝑𝟏 )𝒇 × 𝒔𝒊𝒏𝟓𝟑 − (𝝑𝟐 )𝒇𝒔𝒊𝒏𝟑𝟕

⃗ 𝟏 )𝒇 + 𝟎. 𝟕𝟗(𝝑 ⃗ 𝟐 ) →→ (∗) 𝟓 = 𝟎. 𝟔(𝝑 𝒇

(𝝑𝟐 )𝒇 = (𝝑𝟏 )𝒇 ×

𝒔𝒊𝒏 𝟓𝟑 →→ (∗∗) 𝒔𝒊𝒏 𝟑𝟕

Substitute ** in * : ⃗⃗ 𝟏 )𝒇 + 𝟎. 𝟕𝟗 × ( (𝝑𝟏 )𝒇 × 𝟓 = 𝟎. 𝟔(𝝑

𝒔𝒊𝒏 𝟓𝟑 ) 𝒔𝒊𝒏 𝟑𝟕

⃗ 𝟏 )𝒇 = 𝟑 𝒎/𝒔 (𝝑 ⃗⃗ 𝟐 ) = 𝟑. 𝟗 𝒎/𝒔 → (𝝑 𝒇

‫وليد شاهر أبوخضر‬

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Problems -

Fundamentals and Basics :

12-1 Three balls has the same mass are dropped from altitude ( 𝒉 ) above the ground level . a) Determine each type of collision between the ball and the ground : Ball

Description

Type of collision

𝑨

The ball collides and bounces to an altitude of ( 𝒉 )

𝑨𝒏𝒔 ∶ 𝑬𝒍𝒂𝒔𝒕𝒊𝒄

𝑩

The ball collides and sticks to the ground

𝑨𝒏𝒔 ∶ 𝑷𝒆𝒓𝒇𝒆𝒄𝒕𝒍𝒚 𝑰𝒏𝒆𝒍𝒂𝒔𝒕𝒊𝒄

𝑪

The ball collides and bounces to an altitude of ( )

𝒉 𝟐

b) Which of these balls have done the least work ?

𝑨𝒏𝒔 ∶ 𝑰𝒏𝒆𝒍𝒂𝒔𝒕𝒊𝒄

𝑨𝒏𝒔 ∶ 𝑨

12-2 A moving ball has an elastic head on collision with a second ball of mass ( 𝟎. 𝟕𝟕 𝒌𝒈 ) that is initially at rest , The second ball bounces with one third the original speed of the first ball , Find the following : a) The ratio of the final speed of the first ball to that of the second ball . b) The mass in ( 𝒌𝒈 ) of the first ball . 𝑨𝒏𝒔 ∶ 𝟎. 𝟏𝟓 𝒌𝒈

𝑨𝒏𝒔: 𝟐

12-3 If two symmetrical particles has the same mass collides in a two dimensional collision , where one is moving towards the other resting object , determine the angle between their final momentum vectors ( after collision) . 𝑨𝒏𝒔 ∶ 𝟗𝟎°

12-4 Refer to the motion of two block along a frictionless level track . An initially moving block ( 𝒎𝟏 ) collides and sticks to an initially stationary block of mass ( 𝒎𝟐 = 𝟗 𝒎𝟏 ) , What fraction of the initial kinetic energy of the system is converted into other forms ( heat , sound ,….. ect ) as a result of collision ? 𝟏%

‫وليد شاهر أبوخضر‬

 𝟗𝟎 %

 𝟗𝟗 %

 𝟏𝟎 %

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Higher Level :

12.1 A ball of mass ( 𝟎. 𝟐 𝒌𝒈 ) moves with a velocity of ( 𝟏. 𝟓 𝒊̂ ) 𝒎/𝒔 , meets a second all with a velocity of ( −𝟎. 𝟒 𝒊̂ ) 𝒎/𝒔 in a head on collision . Determine the total impulse done in the collision known that the second ball moved away with ( 𝟎. 𝟗 𝒊̂ ) 𝒎/𝒔 . 𝑨𝒏𝒔 ∶ 𝒛𝒆𝒓𝒐

12.2 Two billiard balls with the same mass ( 𝒂 & 𝒃 ) , collides elastically in a two dimensional collision as shown in the figure , If ball ( 𝒃 ) was at rest. What is the perfect fit of vectors that represents the final momentum of both balls assuming that the initial momentum vector of ball ( 𝒂 ) is represented as { } ? 



𝑝𝑓𝑏 𝜗𝑓𝑎

𝑝𝑓𝑏

 

𝑝𝑖𝑎

𝑝𝑓𝑎

𝑝𝑓𝑎

𝑝𝑓𝑏

𝑝𝑓𝑏

𝑝𝑓𝑎

𝜗𝑖𝑏 = 0

𝜗𝑖𝑎

12.3 The diagram to the right shows the ( velocity – time ) graph for two masses ( 𝑹 & 𝑺 ) that collided elastically , Which of the following statements is wrong ?  The velocities of ( 𝑹 & 𝑺 ) were equal at the mid-time of the collision .  The mass of ( 𝑺 ) was greater than the mass of ( 𝑹 ) .  ( 𝑹 & 𝑺 ) moved in the same direction after the collision .  None of these

‫وليد شاهر أبوخضر‬

𝜗𝑓𝑏

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Finally : Some figures , graphs , problems and paragraphs are copied from some references and books : -

Physics for Scientists and Engineers with Modern Physics - 8th /6th Edition - Serway_Jewett Physics – Harcourt –geoprojects Free high school science texts – FHSST authors

Copying this textbook in other author’s or person’s name without a personal approval from the writer himself is considered a crime and will be severely punished according to law . TM

This work is published on www.academia.edu , for the latest work , papers , researches and summaries follow the author on : https://djasljsa.academia.edu/Kunka/Physics-(-Classical-Mechanics-) Waleed Shaher Abu Khader .

‫وليد شاهر أبوخضر‬

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