When a golfer tees off, the head of her golf club which has a mass of 152 g is traveling 44.8 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 27.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.

56.5 m/s²

Explanation:

From the law of conservation of momentum,

mu+m'u' = mv+m'v' Equation 1

Where m = mass of the golf club, u = initial velocity of the golf club, m' = mass of the golf ball, u' = initial velocity of the golf ball, v = final velocity of the golf club, v' = final velocity of the golf ball.

From the question,

The golf ball is at rest, Hence u' = 0 m/s

mu = mv+m'v'

Make v' the subject of the equation

v' = (mu-mv)/m' Equation 2

Given: m = 152 g = 0.152 kg, u = 44.8 m/s, v = 27.7 m/s, m' = 46 g = 0.046 kg.

Substitute into equation 2

v' = (0.152×44.8+0.152×27.7)/0.046

v' = (6.8096-4.2104)/0.046

v' = 2.5992/0.046

v' = 56.5 m/s²

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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.