Question

The velocity of a ball changes from <8, -7, 0> m/s to <7.95, -7.24, 0> m/s in 0.02 s, due to the gravitational attraction of the Earth and to air resistance. The mass of the ball is 250 grams. (a) What is the acceleration of the ball? (b) What is the rate of change of momentum of the ball? (c) What is the net force acting on the ball?

Please explain. Thank you.

Answer #1

here,

initial speed,u = ( 8 , -7 , 0) m/s

final speed , v = ( 7.95 , - 7.24 , 0) m/s

time taken , t = 0.02 s

a)

the accelration , a = (v-u)/t

a = ( - 0.05 , - 0.24 ,0)/0.02 m/s^2

a = ( - 2.5 , - 12 , 0) m/s^2

b)

the rate of change of momentum of the ball , force = m * (v -u) /t

F = ( - 0.63 , - 3 , 0) N

c)

the net force acting on the ball , |F| = sqrt(0.63^2 + 3^2)

|F| = 3.1 N

direction , theta = arctan(3/0.63)

theta = 258.14 degree counterclockwise from +x axis

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