A particle of mass 100 g moving at an initial speed u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.2 J after the collision, what could be the minimum and the maximum value of u.
Answers
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ANSWER::
Final Kinetic Energy = 0.2 J
Initial Kinetic Energy = (1/2) mV₁² + 0 = (1/2) x 0.1 u² = 0.05 u²
mv₁ = mv₂ = mu
v₁ = Final velocity of 1st block
v₂ = Final velocity of 2nd block
v₁ + v₂ = u Equation - 1
(v₁ - v₂) + L(a₁ - u₂) = 0
La = v₂ - v₁ Equation - 2
u₂ = 0 u₁ = u
Adding Equation - 1 and Equation - 2
2v₂ = (1 + L) u
v₂ = (u/2)(1 + L)
v₁ = u - u/2 - uL/2
v₁ = u(1 - L)/2
Given :-
(1/2) mv₁² + (1/2) mv₂² = 0.2
v₁² + v₂² = 4
[u²(1 - L)²] / 4 + [u²(1 + L)²] / 4 = 4
[u²(1 + L)²] / 2
u² = 8/(1 + L²)
For maximum value of u , denominator should be minimum,
L = 0
u² = 8
u = 2√2 m/s
For minimum value of u , denominator should be maximum,
L = 1
u² = 4
u = 2 m/s
Hope it helps!
