Question

mass A, magnitude 16 is initially IX m/s moving 16 m/'s in the -x-direction. Assurning that the collision is elastic, calculate the final velocity (in m collision is mass e, magnitude 32 velocity C of both masses after the collision m/s) of b collision direction

Answer 1

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