Question

A 2100 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)

Answer 1

Answer:

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

$\overrightarrow{u}=8.89 \widehat{j}$

$\overrightarrow{v}=12.5 \widehat{i}$

Change in momentum = m (v - u)

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

Magnitude of change in momentum

$p = \sqrt{26250^{2}+18669^{2}}$

p = 32211.71 kg m/s

(c) $tan\theta =\frac{-8.89}{12.5}$

θ = 35.4 degree