Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of (20.0m/s)ı^−(4.0m/s)ȷ^(20.0m/s)ı^−( 4.0m/s)ȷ^. During the 3.00 ms that the racket and ball are in contact, the net force on the ball is constant and equal to −(380N)ı^+(110N)ȷ^−(380N)ı^+(110N)ȷ ^. What are the x- and y-components (a) of the impulse of the net force applied to the ball; (b) of the final velocity of the ball?

(a) Jx = -1.14Ns, Jy = 110×3×10-³ = 0.330Ns (b) V = (0m/s)ı^−(1.79m/s)ȷ^

Explanation:

Given

W = 0.56N = mg

m = 0.56/g = 0.56/9.8 = 0.057kg

t = 3.00ms = 3.00×10-³s

Impulse is a vector quantity so we would treat it as such

We have been given the force and velocity in their component forms so to get the impulse from these quantities, we pick the respective component for the quantity we want to calculate and do the necessary calculation. The masses are scalar quantities and so do not affect the signs used in the calculations whether positive or negative. So we have that

u = (20.0m/s)ı^−(4.0m/s)ȷ^

ux = 20m/s

uy = – 4.0m/s

F = – (380N)ı^+(110N)ȷ^

Fx = –380N

Fy = 110N

J = impulse = force × time = F×t

So Jx = Fx ×t

Jy = Fy×t

Jx = –380×3×10-³ = -1.14Ns

Jy = 110×3×10-³ = 0.330Ns

Impulse also equals the change in momentum of the body. So

J = m(v–u)

J/m = v – u

V= J/m + u

Vx = Jx/m + ux

Vx = –1.14/0.057 + 20

Vx = -20 + 20 = 0m/s

Vx = 0m/s

Vy= Jy/m + uy

Vy= 0.33/0.057 + (-4.0)

Vy= 5.79 + (-4.0) = 1.79m/s

V = (0m/s)ı^−(1.79m/s)ȷ^

with 5 seconds, no, but over time if the police car stays at 65m/s ad the speeding car stays at 60, then obviously yes, but the answer to your question would be, no.

hope this : )

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step-by-step explanation: