A 1460 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 18.0 m/s. They stick together. In what direction and with what speed do they move after the collision? Answer in degrees north of east
AND
the speed after the collision in m/s​

Solution given:

North car

mass[m1]=1460kg

velocity[u1]=27 m/s

mass[m2]=2165kg

velocity [u2]=18m/s

let v be velocity after collision

we have

From the principle of conservation of linear momentum

m1u1+m2u2=(m1+m2)v

1460*27+2165*18=(1460+2165)v

v=

v=21.6m/s

the speed after the collision in 21.6 m/s.

For angle.

Tan angle =

Tan angle =

Tan angle=327.74

angle=Tan-¹(327.74)=89.82=90°

in degrees north of east is 90°

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.