Question

A 2400 kg truck traveling north at 36 km/h turns east and accelerates to 55 km/h. (a) What is the change in the truck's kinetic energy? (b) What is the magnitude of the change in its momentum?

(a)Change in kinetic energy[K.E] is the kinetic energy final-kinetic energy initial.Where '-' is the minus sign.

K.E=(1/2)mv^2

Change in K.E=(1/2)(2400)(15.2778)2-(1/2)(2400)(10)2 {the velocities are converted to meter per second}

Change in K.E is 160093.4074 J

(b)Linear Momentum is the product of the mass and the velocity.

Change in linear momentum would be the final momentum-initial momentum.

Change in linear momentum =2400(15.2778i -10j) [i - a unit vector in the positive x direction(towards east) and j - a unit vector in the positive y direction(towards north)]

Magnitude of change in momentum is [ { (2400)^2 } { (15.2778)^2 + (10)^2 } ]^(1/2)
=>43822.92044 kgm/s = 43823 kgm/s (approx)

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