A particle of mass 100 g moving at an initial speed u collides with an

1.	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.
Answer» THANKS for asking the QUESTION! 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!

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.

Answer»

THANKS for asking the QUESTION!

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!