Solution) MA = 2MB
UA = 16 m/s in positive X direction can be taken as (i) in notation.
UA = 16(i) m/s
UB = 16 m/s in negative X direction can be taken as (- i ) in notation.
UB = 16(-i) m/s
VA = ?
VB = ?
Elastic collision both kinetic energy and momentum is conserved.
Applying conservation of momentum
(MA)(UA) + (MB)(UB) = (MA)(VA) + (MB)(VB)
(MA)(UA - VA) = (MB)(VB - UB) ------(1)
Applying conservation of kinetic energy
(1/2)(MA)(UA^2) + (1/2)(MB)(UB^2) = (1/2)(MA)(VA^2) + (1/2)(MB)(VB^2)
(MA)(UA^2 - VA^2) = (MB)(VB^2 - UB^2) -----(2)
A^2 - B^2 = (A+B)(A-B) ----(formula)
(2)/(1) Gives
((MA)(UA + VA)(UA - VA))/((MA)(UA-VA)) =
((MB)(VB - UB)(VB+UB))/(VB - UB)
UA + VA = VB + UB
VB = UA + VA - UB
On substituting in (1)
(MA)(UA - VA) = (MB)(UA+VA-UB-UB)
(MA)(UA - VA) = (MB)(UA+VA-2UB)
(MA)(UA) - (MA)(VA) = (MB)(UA)+(MB)(VA)-(MB)(2UB)
(MB + MA)(VA) = (MA)(UA) - (MB)(UA) + (MB)(2UB)
VA = ((MA - MB)/(MA+MB))(UA) + (2(MB)(UB)/(MA+MB))
VA = ((2MB - MB)/(2MB+MB))(16(i))+(2(MB)(16(-i))/(2MB+MB))
VA = (1/3)(16(i)) + (2/3)(16(-i))
VA = - 5.33(i) m/s
VA = 5.33( - i ) m/s
So velocity magnitude VA = 5.33 m/s and direction is negative X direction .
VB = UA + VA - UB
VB = 16(i) + 5.33(-i) - 16(-i)
VB = 26.67 (i) m/s
Magnitude of velocity is VB = 26.67 m/s
Direction is along positive X direction
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