Item 8 Disk A has a mass of 275 g and is sliding on a smooth horizontal surface

Question

Item 8

Disk A has a mass of 275 g and is sliding on a smooth horizontal surface with an initial velocity (vA)1 = 8 m/s . It makes a direct collision with disk B, which has a mass of 175 g and is originally at rest. Both disks are of the same size and the collision is perfectly elastic (e = 1).

Part A

Determine the velocity of the disk A just after collision.

Express your answer to three significant figures and include the appropriate units.

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Part B

Determine the velocity of the disk B just after collision.

Express your answer to three significant figures and include the appropriate units.

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Part C

Calculate the kinetic energies of the disks before collision T1 and after the collision T2.

Express your answers using three significant figures separated by a comma.

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Item 8

Disk A has a mass of 275 g and is sliding on a smooth horizontal surface with an initial velocity (vA)1 = 8 m/s . It makes a direct collision with disk B, which has a mass of 175 g and is originally at rest. Both disks are of the same size and the collision is perfectly elastic (e = 1).

Part A

Determine the velocity of the disk A just after collision.

Express your answer to three significant figures and include the appropriate units.

(vA)2 =

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Part B

Determine the velocity of the disk B just after collision.

Express your answer to three significant figures and include the appropriate units.

(vB)2 =

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Part C

Calculate the kinetic energies of the disks before collision T1 and after the collision T2.

Express your answers using three significant figures separated by a comma.

T1, T2 =   J  

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Explanation / Answer

a)

Ma = mass of disk A = 275 g

Mb = mass of disk B = 175 g

Vai = velocity of disk A before collision = 8 m/s

Vbi = velocity of disk B before collision = 0 m/s

using conservation of momentum

Ma Vai + Mb Vbi = Ma Vaf + Mb Vbf

(275) (8) + (175) (0) = (275) Vaf + (175) Vbf

Vbf = (2200 - (275) Vaf )/175                       eq-1

using conservation of kinetic energy

Ma V2ai + Mb V2bi = Ma V2af + Mb V2bf

(275) (8)2 + (175) (0)2 = (275) V2af + (175) V2bf

17600 = (275) V2af + (175) ((2200 - (275) Vaf )/175)2

using eq-1

Vaf = 1.78 m/s

b)

Vbf = (2200 - (275) Vaf )/175

Vbf = (2200 - (275) (1.78) )/175 = 9.8 m/s

c)

KEi = kinetic energy before collision = (0.5) (Ma V2ai + Mb V2bi ) = (0.5) ((275) (8)2 + (175) (0)2 )

KEi = 8.8 J

KEf = kinetic energy after collision = (0.5) (Ma V2af + Mb V2bf ) = (0.5) ((0.275) (1.78)2 + (0.175) (9.8)2 )

KEf = 8.8 J

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