Challenge Questions

This week's Challenge Question:

A rocket is about to be launched from earth. Its mass before the launch is 1000 kg and when moving it consumes fuel at a rate of 2.5 kg/sec. If the speed of the exhaust gases relative to the rocket engine is 2500 m/sec, determine its velocity 10 seconds after the launch.

The Answer will be posted right here on CR4 on November 24th.

This week's Challenge Question:

A rescue plane flies at 198 km/hr and at a constant elevation of 500 m trying to rescue a person involved in a boating accident. The pilot releases a rescue capsule so that it hits the water close to the victim. Determine the angle (A, in the figure) of the pilot's line of sight when the release is made.

And the answer is...

As indicated in the given figure, we will put the origin of our coordinate system at the point of release of the capsule. Let's make the time of release as t = 0.

Angle A is given by

Because the capsule, once released, is a projectile, it follows that the horizontal and vertical velocities are independent. The horizontal velocity of the capsule is the same as that of the plane, and it is constant, or its initial velocity is equal to its velocity at any time t. Thus we can express the horizontal velocity by

Or,

Where tf is the time the capsule reaches its destination. The initial velocity is 198 km/h or 55 m/s.

We see now that to determine d we must calculate tf. This can be accomplished if we consider the vertical displacement. For a projectile released at time t = 0, the vertical displacement is given by

At t = tf the displacement is y = -500 m. With g = 9.8 m/s2, using the equation given above we get tf = 10.1 seconds. Then the horizontal distance, d, is calculated as

Finally the angle A is calculated

This month's Challenge Question:

The three apple farmers (Mike, Juliet, and Bob) finished their harvest and reported yields of 314,827 apples, 1,199,533 apples, and 683,786 apples, respectively. Mike mentioned the number of apples he'd have left over if he divided his harvest equally among all the apple dealers in the area. Juliet indicated she would have bought the extra apples so she could divide her supply equally among the dealers, with Bob indicating the same thing. How many apple dealers are there?

The Answer will be posted right here on CR4 on December 2nd. Can't wait that long? Well, check out these weekly challenges from CR4:

Chain Links: CR4 Challenge (10/27/09)

Daughters and Sons: CR4 Challenge (10/20/09)

Chain and Pendant: CR4 Challenge (10/13/09)

This week's Challenge Question:

A customer goes into a jewelers shop and lays 8 chain segments on the counter. The segments are 30, 20, 17, 13, 9, 6, 3, 2 links long long, and each link is 1cm. The customer asks that they be assembled into a continuous chain 100 links long, by cutting and re-joining links as necessary. If it takes 1 minute to cut and open a link, and 2 minutes to weld it closed (ignore the time taken to pick-up, thread or set down the link), what is the least time in which the jeweler can complete the task?

Thanks to SlideRuler for providing us with this Challenge.

And the Answer is...

There are 8 sections. If he simply opened the end link of a section and attaches it to another section, he would have to do this 7 times to make a continuous length taking 21 minutes. If he cuts the 6 link section into separate links, he will have enough cut links to join the 7 remaining sections, taking 18 min. He can do even better, by cutting the 3 and 2 link sections into separate links, and use the 5 cut links to join the 6 remaining sections. Time taken in this case is 15 min.

This week's Challenge Question:

You have three colleagues: Arthur, Bill and Charlie. Each has two (and only two) children. Bill mentioned that today is his daughter's birthday and Charlie brought his daughter to work.

Assuming 50:50 male:female distribution and no further information about your colleagues or their families, what are the probabilities that each has a son?

Thanks to ExPat for providing us with this Challenge.

And the Answer is....

Arthur: 3/4

Bill: 2/3

Charlie: 1/2

Most won't have difficulty with Arthur and Charlie; Bill's case is less intuitive, at least until the problem is broken down.

Explanation:

Designate the two children and Child A and Child B. Each child can be a boy or a girl (B or G). For Arthur, there are four permutations, each with equal probability; three include at least one boy, so the probability that Arthur has a son is 3/4. Child A Child B Has Son?

For Bill, we know that B-B is not an available solution, leaving three permutations, each with equal probability; two include a boy, so the probability that Bill has a son is 2/3. Child A Child B Has Son?

For Charlie, we have already met one child (let's designate her Child A), so solutions B-B and B-G are unavailable, leaving two permutations, each with equal probability; one includes a boy, so the probability that Charlie has a son is 1/2. Child A Child B Has Son?

(We could of course designate the known daughter as Child B, resulting in the same probabilities.)