The track shown in figure (9-E16) is frictionless. The block B of mass 2m is lying at rest and the block A of mass m is pushed along the track with some speed. The collision between A and B is perfectly elastic. With what velocity should the block A be started to get the sleeping man awakened?
Answers
Thanks for asking the question!
ANSWER::
Let velocity of A = u₁
Let final velocity when block reaching at B becomes collision = v₁
Therefore , (1/2) mv₁² - (1/2) mu₁² = mgh
v₁² - u₁² = 2gh
v₁ = √(2gh - u₁²) [Equation 1]
When block B reach at upper man's head , the velocity of B is just 0
For block B ,
(1/2) x 2m x 0² - (1/2) x 2m x v² = mgh
v = √(2gh)
Before collision velocity of block A , u₃ = v₁ u₄ = 0
After collision velocity of block A , v₃ = v (say) v₄ = √(2gh)
Since , it is an elastic collision the momentum and kinetic energy should be conserved.
m x v₁ + 2m x 0 = m x v + 2m x √(2gh)
v₁ - v = 2√(2gh)
And also ,
(1/2) x m x v₁² + (1/2) x 2m x 0² = (1/2) x m x v² + (1/2) x 2m x [√(2gh)]²
v₁² - v² = 2 x √(2gh) x √(2gh) [Equation 2]
Dividing Equation 1 by Equation 2
[(v₁ + v)(v₁ - v)] / (v₁ + v) = [2 x √(2gh) x √(2gh)] / [2 x √(2gh)]
v₁ + v = √(2gh) [Equation 3]
Now , adding Equation 1 and Equation 3
2v₁ = 3 √(2gh)
v₁ = 3√(2gh) / 2 [Equation 4]
But , v₁ = √(2gh + u²) [Equation 5]
Equating Equation 4 and Equation 5
2gh + u² = 9(2gh) / 4
u = 2.5√2gh
So , block will travel with velocity greater than 2.5√2gh so , as to awake man by B.
Hope it helps!
