Question

Hello friends!!!

A 60kg man skating with speed of 10m/s collides with 40 kg skater at rest and they cling to each other find the loss of kinetic energy during the collision .

Answer 1

$\ \bold {here \: is \: your \: answer}$
given :

Mass of the man , m₁ = 60 kg

Speed of the man , v₁ = 10 m/s

Mass of the skater , m₂ = 40 kg

Let velocity of skater after collision = v₂

so......


suppose 60 kg man has mass m1

supposed his velocity be v1

supposed 40 kg man has mass m2

Initial kinetic energy = m1v^2/2
=(60×10×10)/2=3000Joule

According to conservation of momentum,

m1v1 +m2v2 =( m1+m2)V

60×10 + 40×0 =(60+40)V

V=6m/s

Final kinetic energy ={(m1+m2)v^2}/2=(100*6*6)/2= 1800Joule

Therefore loss in kinetic energy = 3000-1800=1200Joule answer.......


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Answer 2

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________

$\bold \green{solution}$ :-

________________

Given :-

Mass of the man = m1 = 60 kg

speed of man V1 = 10 m/s

mass of skater = m2 = 40kg

velocity of skater V2 = 0 ( before collision )

let velocity of system after collision= V' Since,

Using law of momentum of conservation

m1V1 + m2*0 = (m1 + m2 )V' 60 * 10 + 0 = (60 + 40) V'600 = 100V2 V' = 6m/s

•°• loss in K.E = 1/2 *60*(10)² - 1/2* 100*(6)²

=> 3000 - 1800

=> - 1200 J since, loss in kinetic energy = 1200J

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