Question

A block of mass 0.50kg is moving with a speed of 2.00 m/s on a smooth surface . it strikes another mass of 1kg and then they move together as a single body . the energy loss during the collision is ​

Answer 1

Answer:

ANSWER

R.E.F image

By momentum  conservation 

0.5×2=1.5×v

v=32m/sec

Initial Energy  =21×0.5×(2)2

=1 joule .

Final Energy  =21×1.5×(32)2

=1/3 joule .

Loss= Initial-Final

=1−31=32

Answer 2

Explanation:- From law of conservation of momentum , we have

• m₁v₁ + m₂v₂ = (m₁+m₂)v

Given :- m₁ = 0.50kg , V₁ = 2m/s

m₂ = 1kg , v₂ = 0 (at rest )

0.5 × 2 + 1× 0 = 1.5×v

[assumed that 2nd body is at rest ]

• ⟹ v = 2/3

• ∴ ∆K = Kᵢ- Kf
• = 1.5× (2/3)²/2 - 0.5× (2²)/2 = - 2/3J
• =-: 0.67J

So , energy most is 0.66 J . Answer .