Question

a ping pong ball weighing 2 oz. with a contant velocity of V=8j+6k ft/sec. A gust of wind exerts a force of F=.5t i oz. on the ball. Determine the magnitude and direction of th ball after .5 sec

Answer 1

Weight of ball = 2oz

1 oz force = 0.278 N

0.5 oz force = 0.5 * 0.278 = 0.139 N

F = ma

m = 2 oz = 2 * 0.028 kg = 0.056 kg

Force exerted on ball by wind = 0.139 i N

Thus acceleration imparted by the wind a_w = F/m = 0.139 / 0.056 = 2.48 i m/s²

Now in addition to force exerted by wind that is in +x direction, earth's gravity also imparts force on ball g = 9.8 m/s² in -y direction.

Thus total acceleration on ball=a_w - g = 2.48 i -9.8j m/s²

original velocity of ball: u = 8j + 6k ft/s = 8*0.305 j + 6*0.305 k m/s = 2.44 j + 1.83 k m/s

Velocity after t=0.5 s

final velocity v = u + at

v = (2.44 j + 1.83 k) + ((2.48 i -9.8)* 0.5)) = (2.44 j + 1.83 k) + (1.24 i - 4.9j)

or,

v = 1.24 i + (2.44-4.9)j + 1.83k

ANS: v = (1.24 i - 2.46 j + 1.83 k) m/s

Magnitude of final velocity:

v = square root (1.24² + (-2.46)² + 1.83²) = square root (1.538 + 6.052 + 3.35) = 3.31 m/s

Direction of final velocity:

we find direction cosines:

theta_1 = angle final velocity vector makes with respect to x axis.

cos (theta_1) = x / square root (x² + y² + z²) = 1.24/3.31 = 0.375

or, theta_1 = cos^-1 (0.375) = 67.98 deg

Similarily we find angle that v makes with y axis:

cos (theta_2) = y / square root (x² + y² + z²) = -2.46/3.31 = -0.74

or, theta_2 = cos^-1 (-0.74) = 137.8 deg

And the vector v makes with z axis:

cos (theta_3) = z / square root (x² + y² + z²) = 1.83/3.31 = 0.553

or, theta_3 = cos^-1 (0.553) = 56.43 deg

Thus the vecor makes:

with x axis: 67.98 deg

with y axis: 137.8 deg

with z axis: 56.43 deg