Figure (9-E22) shows a small body of mass m placed over a larger mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed v and the system is left to itself. Assume that all the surfaces are frictionless.
(a) Find the speed of the larger block when the smaller block is sliding on the vertical part.
(b) Find the speed of the smaller mass when it breaks off the larger mass at height h.
(c) Find the maximum height (from the ground) that the smaller mass ascends.
(d) Show that the smaller mass will again land on the bigger one. Find the distance traversed by the bigger block during the time when the smaller block was in its flight under gravity.
Answers
Thanks for asking the question!
ANSWER::
Mass m is given velocity v over the larger mass M.
(a) When smaller block is travelling on vertical part , then let the velocity of bigger block be v₁ towards left.
Now , law of conservation of momentum , ( in horizontal direction )
mv = ( M + m) v₁
v₁ = mv / (M + m)
(b) When smaller block breaks off , then let the resultant velocity is v₂
Now , law of conservation of energy ,
(1/2) mv² = (1/2) Mv₁² + (1/2) mv₂² + mgh
v₂² = v² - (Mv₁² / m) - 2gh [ Equation 1 ]
v₂² = v² [ 1 - (M/m) x {m² / (M + m)²} ] - 2gh
v₂ = √[ {(m² + Mm + m²)v² / (M + m)²} - 2gh]
(c) Vertical component of velocity v₂ of mass m is given by,
v₃² = v₂² - v₁²
v₃² = {(m² + Mm + m²)v² / (M + m)²} - 2gh - [m²v² / (M + m)²] {v₁ = mv / (M + v)}
v₃² = [(M² + Mm + m² - m²)v² / (M + m)²] - 2gh
v₃² = [Mv² / (M + m)] - 2gh [Equation 2]
To find maximum height ( from ground ) , let us assume the body rises to a height h , over and above h
(1/2) mv₃² = mgh₁
h₁ = v₃² / 2g [ Equation 3 ]
Total height = h + h₁ = h + v₃² / 2g = h + [mv² / (M + m)2g] - h
Now , from Equation 2 and Equation 3
H = mv² / (M + m)2g
(d)Smaller mass has also got a horizontal component of velocity v₁ at time it breaks off from M which has a velocity v₁ and the block m will again land on the block M
Let us find out time of flight of block m after it breaks off.
During upward motion from B to C
0 = v₃ - gt₁
t₁ = v³/g
t₁ = (1/g) √[ {Mv² / (M + m)} - 2gh ] [ Equation 4 {from Equation 2}]
So , the time for which smaller block was in its flight is given by,
T = 2t¹ = (2/g) √[ {Mv² - 2 (M + m)gh} / (M + m)]
And the distance travelled by bigger block during this time is,
S = v₁T = [mv / (M+m)] x (2/g) √[ {Mv² - 2 (M + m)gh} / (M + m)]
S = 2mv√[Mv² - 2(M + m)gh] / g(M+m)^(3/2)
Hope it helps!
