Question

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

Answer 1

let 60 kg man has mass m1
let his velocity be v1
let 40 kg man has mass m2
Initial kinetic energy = m1v^2/2
=(60×10×10)/2=3000J
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= 1800J.
Therefore loss in kinetic energy = 3000-1800=1200J

Answer 2

Thanks for asking the question!

ANSWER::

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

Therefore , 60 x 10 + 0 = 100 x v₂

v₂ = 6 m/s

Loss in Kinetic Energy = (1/2)m₁v₁² - (1/2)m₂v₂² = (1/2)x60x10² - (1/2)x100x6²

                                                                                 = 3000-1800=1200 J

Hope it helps!