Question

A 14,000 kg tractor traveling north at 21 km/h turns west and travels at 26 km/h . Calculate the

a) change in the tractor's kinetic energy. I think it's 1.27 x 10^5 J

b) linear momentum (magnitude and direction). I think I got the magnitude?? 8.22 x 10^5 kg m/s but how do I get direction?

Answer 1

(a) v_i = 21 km/h = 21 * (5/18) m/s = 5.83 m/s

v_f = 26 km/h = 26*(5/18) m/s = 7.22 m/s

Change in KE = KE_f - KE_i

= 1/2*m*(v_f² - v_i²)

= 1/2*14000*(7.22² - 5.83²)

= 1.27 x 10⁵ J

(b)

Draw the final vector in the negative direction (west) and label it with -7.22, then I flip the direction of the north vector (v initial) making it south to subtract from the v final. I get a vector pointing west, plus a vector pointing south. I draw an arrow from the tail of first to tip of last. I then have a right angle, do pythagorean theroem to find the magnitude.

(since for vectors, A –B = A + (-B) )

Change in momentum

= m |v_f - v_i|

=14000 * sqrt((-7.22)² + (-5.83)²) (by Pythagorus theorem, since North and East are perpendicular)

= 1.3 x 10⁵ kg m/s

Angle(direction)

= arctan(5.83/7.22) below -x

= 38.9 degrees below -x axis