Two balls that their mass is m1=m and m2=2m are connected to the ceiling through
ID: 1780593 • Letter: T
Question
Two balls that their mass is m1=m and m2=2m are connected to the ceiling through non-mass wires. The length of each wire is L The light ball is brought to the position where the wire connected to the ball is stretched horizontally and released after the collision. The two balls continue to move in the same direction and the heavy ball rises above the collision point 4 of the light ball. A. What is the velocity of the light ball before collision? B. The velocity of each ball after the collision? C. what is the maximum height each ball rise? D.What is the percentage of the energy that has become heat?
Explanation / Answer
(A) from energy conservation,
PEi + Kei = PEf + KEf
m g L + 0 = 0 + m v^2 / 2
v = sqrt(2 g L) .....Speed of light ball
speed of heavy ball = 0 (it was at rest)
(B) suppose light rises by the height h then heavy ball height is 4h.
v_h = sqrt(2 g (4h)) = 2 sqrt(2 g h)
v_L = sqrt(2 g h)
Applying momentum conservation,
m sqrt(2 g L ) = m sqrt(2 g h) + 2 m 2 sqrt(2 g L)
2 g L = 25 x 2 g h
h = L/25
of heavy ball =2 sqrt(2 g L) / 5
of light ball = sqrt(2 g L ) / 5
(C) maximum heights, of heavy ball = 4 L / 25
of light ball = L/25
(D) Ki = m g L
Kf = m g L / 25 + 8 m g L / 25 = 9 m g L / 25
energy lost = Ki - Ki = 16 m g L / 25