Abox of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to a waiting truck. it happens a box of m2 = 40.0 kg was left at the bottom of the chute. what is the impulse acting on m2 during the collision between the two boxes?

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass just before the collision

Therefore the momentum just before collision would be

Momentum after the collision

Since the momentum is conserved we have that

The velocity of mass after the collision is given by

Therefore the change in momentum of mass 2 is

Therefore the impulse acting on m2 during the collision between the two boxes is

their crystal patterns have been disrupted by stress

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.