Question

A 143-g baseball is flying through the air with a speed of 180 km/hr just after it is hit by a bat. If its velocity is at an angle of 37º above the horizontal and 22º north of east, find its momentum in unit-vector notation. Define east as x and north as y

Answer 1

To solve the problem it is necessary to apply the concepts related to the conservation of linear Moment, that is to say

$P=mv$

Where,

m = Mass

v = Velocity

P = Linear momentum

For the given data we have to:

$v_1 = 180\frac{km}{h}(\frac{1h}{3600s})(\frac{1000m}{1km})$

$v_1 = 50m/s$

The components of this force would be given by,

$v_z = 50*sin37 = 30.09m/s$

$v_{x,y} = 50*cos37=39.93m/s \rightarrw$According to the definition given at the end of the problem, this component corresponds to that expressed for x and y.

Applying the previous equation we have,

$\vec{P} = m(39.93cos\theta \hat{i}+39.93sin\theta \hat{j}-30.09\hat{k})$

Note: The component at this direction must also decomposed

The mass is 143g=0.143kg, then:

$\vec{P} = (0.143)(39.93cos(22)\hat{i}+39.93sin(22) \hat{j}-30.09\hat{k})$

Therefore the final vector is:

$\vec{P} = 5.29\hat{i}+2.138\hat{j}-4.29\hat{k})$