Question

The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 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 120 grams.

(a) What is the acceleration of the ball? a with arrow = Correct: Your answer is correct. (m/s)/s

(b) What is the rate of change of momentum of the ball? dp with arrow/dt = Incorrect: Your answer is incorrect. (kg · m/s)/s

(c) What is the net force acting on the ball? F with arrownet = N

Answer 1

a)
`a=(-2,-6,0)m/s^2`, with a magnitude of
`6.3m/s^2`

b)
`(\Delta p)/(\Delta t)=(-0.24,-0.72,0)Kgm/s^2`, with a magnitude of
`0.76Kgm/s^2`

c)
`F=(-0.24,-0.72,0)N`, with a magnitude of
`0.76N`

Step-by-step explanation:

We have:

`v_(ix)=9m/s, v_(iy)=-6m/s, v_(iz)=0m/s\\v_(fx)=8.96m/s, v_(fy)=-6.12m/s, v_(fz)=0m/s\\t=0.02s, m=0.12Kg`

We can calculate each component of the acceleration using its definition
`a=(\Delta v)/(\Delta t)`

`a_x=(v_(fx)-v_(ix))/(t) = ((8.96m/s)-(9m/s))/(0.02s) =-2m/s^2\\a_y=(v_(fy)-v_(iy))/(t) = ((-6.12m/s)-(-6m/s))/(0.02s) =-6m/s^2\\a_y=(v_(fz)-v_(iz))/(t) = ((0m/s)-(0m/s))/(0.02s) =0m/s^2\\`

The rate of change of momentum of the ball is
`(\Delta p)/(\Delta t) = (\Delta mv)/(\Delta t) = (m\Delta v)/(\Delta t) = ma`

So for each coordinate:

`(\Delta p_x)/(\Delta t)=-0.24Kgm/s^2\\(\Delta p_y)/(\Delta t)=-0.72Kgm/s^2\\(\Delta p_z)/(\Delta t)=0Kgm/s^2`

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

`a=√(a_x+a_y+a_z) =6.3m/s^2\\F=ma=(\Delta p)/(\Delta t)=0.76N`

(N and
`Kgm/s^2` are the same unit).