Answer
given,
mass of truck = 2100 Kg
velocity in north direction = 41 km/h
= 41 x 0.278 m/s
= 11.398 m/s
velocity in north direction = 51 km/h
= 51 x 0.278 m/s
= 14.178 m/s
a) Change in K.E
= [tex]\dfrac{1}{2}m(V_f^2-V_i^2)[/tex]
= [tex]\dfrac{1}{2}\times 2100\times (14.178^2-11.398^2)[/tex]
= 7.453 x 10⁴ J
b) Change in momentum
[tex]\Delta P = m(v_f-V_1)[/tex]
[tex]\Delta P = 2100\times (14.178 - 11.398)[/tex]
[tex]\Delta P = 5838\ kg.m/s[/tex]
c) Direction of momentum
[tex]\theta = tan^{-1}(\dfrac{P_y}{P_x})[/tex]
[tex]\theta = tan^{-1}(\dfrac{11.398}{14.178})[/tex]
[tex]\theta = 38.796^0[/tex]
