A 15 kg block is sliding along a frictionless surface and strikes a 10 kg ball at rest. What is the collision of the blocks after the collision if it is an inelastic collision?

   v = 0.6 v₁

Explanation:

This is an exercise in collisions, let's start by defining a system formed by the two bodies, so that the forces during the collisions have been internal and the momentum is preserved.

Instant starts. Before the crash

       p₀ = M v₁ + m 0

Final moment. After the crash

        = (M + m) v

how momentum is conserved

       p₀ = p_{f}

       M v₁ = (M + m) v

       v =

let's calculate

       v =

        v = 0.6 v₁

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.

i would say