Problem 9.106 Part A Calculate if car A approaches at 4.70 m/s and car B is movi

Question

Problem 9.106 Part A Calculate if car A approaches at 4.70 m/s and car B is moving at 3.85 m/s, their velocities after the collision Enter your answers numerically separated by a comma Two bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear, see the figure.(Figure 1) Car A has a mass of 425 kg and car B 500 kg, owing to differences in passenger mass m s Request Answer Figure 1 of1 Part B Calculate if car A approaches at 4.70 m/s and car B is moving at 3.85 m/s, the change in momentum of each. Enter your answers numerically separated by a comma. IM m/5

Explanation / Answer

In a perfectly elastic collision

Using momentum conservation

Pi = Pf

m1V1i + m2V2i = m1V1f + m2*V2f

given that m1 = 425 kg & m2 = 500 kg

V1fi = 4.70 m/sec & V2i = 3.85 m/sec

425*4.70 + 500*3.85 = 425*V1f + 500*V2f

425*V1f + 500*V2f = 3922.5

Now In elastic collisions,

V1f - V2f = V2i - V1i

V1f - V2f = 3.85 - 4.70

V1f - V2f = -0.85

Now Solving both equation

Multiply equation 2 with 500 and Add both of them

925*V1f = 3922.5 - 500*0.85

V1f = (3922.5 - 500*0.85)/925 = 3.781 m/sec (Va)

V2f = V1f + 0.85

V2f = 3.781 + 0.85 = 4.631 m/sec (Vb)

Part B

dPa = m1*V1f - m1*V1i = 425*(3.781 - 4.70) = -390.5 kg-m/sec

dPb = m2*V2f - m2*V2i = 500*(4.631 - 3.85) = 390.5 kg-m/sec

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