Solution For this problem, note that v 0 and use conservation of momentum. Thus,

Question

Solution For this problem, note that v 0 and use conservation of momentum. Thus, (8.37) P1 p 1 p 2 or (8.38) m 1 Vi 2 2. Using conservation of internal kinetic energy and that v -0, (8.39) 2 m 1 V m 1 V 2 2 2 Solving the first equation (momentum equation) for v'2, we obtain (8.40) Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable v eaving only v 1 as an unknown (the algebra is left as an exercise for the reade). There are two solutions to any quadratic equation; in this example, they are (8.41) 4.00 m/s v 1 and (8.42) 1 3.00 m/s

Explanation / Answer

m1 = 0.5 kg, m2 = 3.5 kg

u1 = 4, u2 = 0

let the final velocities be V1 and V2

now, 0.5*4 = 0.5V1 + 3.5V2

5V1 + 35V2 = 20

V1 + 7V2 = 4 ---------1

Energy equation gives

8V1^2 = 0.5V1^2 + 3.5V2^2

V1^2 + 7V2^2 = 16

and V2 = 1/7 *(4 - V1)

V1^2 + 1/7 (16 + V1^2 - 8V1) = 16

7V1^2 + 16 + V1^2 - 8V1 = 112

8V1^2 - 8V1 - 96 = 0

V1^2 - V1 - 12 = 0

solving this equation gives

V = 4 m/s and -3 m/s

solving this equation gives

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