A2100 kg truck traveling north at 32 km/h turns east and accelerates to 45 km/h.
(a) what is the change in the truck's kinetic energy? j
(b) what is the magnitude of the change in the linear momentum of the truck? kg·m/s
(c) what is the direction of the change in the linear momentum of the truck? ° (measured clockwise from east)

Explanation:

mass of truck, m = 2100 kg

initial velocity, u = 32 km/h north = 8.89 m/s

final velocity, v = 45 km/h east = 12.5 m/s

(a) Initial kinetic energy, Ki = 1/2 m u²

Ki = 0.5 x 2100 x 8.89 x 8.89 = 82983.71 J

Final kinetic energy, Kf = 1/2 mv²

Kf = 0.5 x 2100 x 12.5 x 12.5 = 164062.5 J

(b)

Write the velocities in vector form

Change in momentum = m (v - u)

p = 2100 (12.5 i - 8.89 j ) = 26250 i - 18669 j

Magnitude of change in momentum

p = 32211.71 kg m/s

(c)

θ = 35.4 degree

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s