An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. the air film under the puck is 0.1 mm thick. calculate the time required after impact for a puck to lose 10% of its initial speed. assume air is at 15o c and has a dynamic viscosity of 1.75´10-5 n×s/m2 .

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × m

dynamic viscosity = 1.75 × Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ

so

= µ     ............1

put here value

= 1.75× ×

= 0.175 v

and

area between air and puck is given by

Area =

area  =  

area = 7.85 × m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 ×

force = 1.374 × v    

and now apply newton second law

force = mass × acceleration

- force =

- 1.374 × v =

t =  

time = 2.18

so time required after impact for a puck is 2.18 seconds

momentum of soft ball three times less than baskey ball

c) up

hope this