Question

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

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?

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

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

Change in KE = KEf - KEi

= 1/2*m*(vf2 - vi2)

= 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 |vf - vi|

=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

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