Question

#19 A 2100 kg truck travel north at 41 km/h turns east and accelerates to 51 km/h.

(a) What is the change in the truck’s kinetic energy? What are the
(b) magnitude
(c) direction of the change in its momentum?

Answer 1

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

=
`(1)/(2)m(V_f^2-V_i^2)`

=
`(1)/(2)* 2100* (14.178^2-11.398^2)`

= 7.453 x 10⁴ J

b) Change in momentum

`\Delta P = m(v_f-V_1)`

`\Delta P = 2100* (14.178 - 11.398)`

`\Delta P = 5838\ kg.m/s`

c) Direction of momentum

`\theta = tan^(-1)((P_y)/(P_x))`

`\theta = tan^(-1)((11.398)/(14.178))`

`\theta = 38.796^0`