Momentum and Energy in Elastic and Inelastic Collisions

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Presentation transcript:

Momentum and Energy in Elastic and Inelastic Collisions Collisions Revisited! Momentum and Energy in Elastic and Inelastic Collisions

Reminder: Conservation of Momentum In a closed system, the total momentum of all objects in the system will remain constant In other words…the total momentum in the system right before an “event” will be equal to the total momentum right after the “event”

Reminder: Kinetic Energy (EK) The energy a body has because it is moving Units: Joules (J) http://news.nationalgeographic.com/news/2012/08/120802-cheetah-sarah-cincinnati-zoo-fastest-record-science-usain-bolt-olympics/# Cheetah Breaks Speed Record - Beats Usain Bolt by Seconds

Types of Collisions and Conservation Kinetic Energy Momentum Elastic: Objects bounce off each other undamaged Conserved Conserved Inelastic: Objects impact and separate, but there has been damage done to each NOT conserved Conserved Objects impact and STICK TOGETHER as one larger mass with the same, slower velocity Perfectly Inelastic: NOT conserved Conserved

Types of Collisions and Conservation Kinetic Energy Momentum Elastic: 𝑚1𝒗𝟏+𝑚2𝒗𝟐 = 𝑚1𝒗𝟏’+𝑚2𝒗𝟐’ 1 2 𝑚 1 𝑣1 2 + 1 2 𝑚 2 𝑣2 2 = 1 2 𝑚 1 𝑣′1 2 + 1 2 𝑚 2 𝑣′2 2 Conserved Conserved Inelastic: 𝑚1𝒗𝟏+𝑚2𝒗𝟐 = 𝑚1𝒗𝟏’+𝑚2𝒗𝟐’ NOT conserved Conserved Perfectly Inelastic: 𝑚1𝒗𝟏+𝑚2𝒗𝟐 =(𝑚1+𝑚2)𝒗’ NOT conserved Conserved

Example problem 1: A freight train is being assembled in a switching yard. Boxcar #1 has a mass of 6.5 x 104 kg, and Boxcar #2 has a mass of 9.2 x 104 kg. If car 1 is moving with a velocity of +0.80 m/s, and car 2 hits it from behind with a velocity of +1.2 m/s, with what velocity will the two cars move with together after coupling?

Example problem 1: solution 𝒑 𝒃𝒆𝒇𝒐𝒓𝒆 = 𝒑 ′ 𝒂𝒇𝒕𝒆𝒓 𝑚1𝒗𝟏+𝑚2𝒗𝟐 = 𝑚1+𝑚2 𝒗’ (6.5× 10 4 kg)(0.80 m s )+(9.2× 10 4 kg)(1.2 m s )= 6.5× 10 4 kg+9.2× 10 4 kg 𝒗’ (52000 kg∙m s )+(110400 kg∙m s )= 157000 kg 𝒗’ 𝒗’ = (162400 kg∙m s ) 157000 kg =𝟏.𝟎 𝒎 𝒔

Example Problem 2: Starting from rest, two ice skaters push off against each other on smooth, level ice. Skater 1 has a mass of 54 kg and moves away from her friend (mass = 88 kg) with a velocity of 2.5 m/s. Find the “recoil” velocity of the second skater.

Example Problem 2: Solution 𝒑 𝒃𝒆𝒇𝒐𝒓𝒆 = 𝒑 ′ 𝒂𝒇𝒕𝒆𝒓 𝑚1𝒗𝟏+𝑚2𝒗𝟐 =𝑚1𝒗′𝟏+𝑚2𝒗′𝟐 0+0=(54 kg)(2.5 m s )+(88 kg)𝒗′𝟐 𝒗’2= (−135 kg∙m s ) 88 kg =−𝟏.𝟓 𝒎 𝒔