A truck of mass 2000kg travels with speed 20 m/s at 60degree North of East and c

Question

A truck of mass 2000kg travels with speed 20 m/s at 60degree North of East and collides with a car of mass 1000kg travelling at 28.285 m/s at 45degree North of West. The two vehicles stick together and continue to move as a single unit with no friction. Use the principle of conservation of momentum to calculate: a) The speed with which the resulting wreck now moves b) The direction in which it moves. c) Is energy conserved? d) Calculate the difference (if any) between the initial and final energies of the car-truck system.

Explanation / Answer

here,

m1 = 2000 kg

m2 = 1000 kg

u1 = 20 * (cos(60) i + sin(60) j )

u1 = (10 i + 17.32 j)m/s

u2 = 28.85 * ( - cos(45) i + sin(45) j)

u2 = - 20.4 m/s i + 20.4 m/s j

let the final velocity be v

using conservation of momentum

m1 * u1 + m2*u2 = ( m1 + m2) * v

2000 * (10 i + 17.32 j ) + 1000 * (- 20.4 i + 20.4 j ) = 3000 * v

v = - 0.13 i + 18.35 j

|v| = sqrt( 0.13^2 + 18.35^2)

|v| = 18.35 m/s

theta = arctan(0.13/18.35)

theta = 0.41 degree

(a)

the resulting wreek moves with a speed of 18.35 m/s

(b)

the angle is 0.41 degree west of north

(c)

No , the energy is not conserved

(d)

the change in energy , KE = 0.5 * ( m1 + m2) * v^2 - 0.5 * m1*u1^2 - 0.5* m2*u2^2

KE = 0.5 * ( 3000) * 18.35^2 - 0.5 * 2000*20^2 - 0.5 *1000 28.285^2

KE = - 2.87 * 10^5 J

the change in kinetic energy is - 2.87 * 10^5 J

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