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_{f}^{2} - v_{i}^{2})

= 1/2*14000*(7.22^{2} - 5.83^{2})

= **1.27 x
10 ^{5} 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
10 ^{5} kg m/s**

Angle(direction)

= arctan(5.83/7.22) below -x

= **38.9 degrees
below -x axis**

A 1990 kg truck traveling north at 45 km/h turns east and
accelerates to 56 km/h. (a) What is the change in the truck's
kinetic energy? (b) What is the magnitude of the change in its
momentum?

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