Our hapless hero has once again gotten himself into trouble on the frictionless

Question

Our hapless hero has once again gotten himself into trouble on the frictionless ice. This time he has somehow acquired a constant velocity towards a region of thin ice; if he cannot find a way to stop he will slide into the region of thin ice, break through the ice, and fall into the freezing water. Our hero finds himself going North at 38.0 cm/s. Luckily, he has his physics book with him as he is sliding on the ice. (a) How can our hapless hero use his physics book to stop his northward motion and keep from falling into the freezing water? Throw the physics book towards the East. Throw the physics book towards the North. Throw the physics book straight Down onto the ice. Throw the physics book towards the West. Throw the physics book towards the South. Throw the physics book straight Up. (b) The mass of our hero is 60.0 kg, and the mass of his physics book is 2.4 kg. What is the total momentum of the system of particles consisting of the book and our hero combined as they are slipping across the ice together BEFORE our hero throws his physics book? Give your answer in kg·m/s. kg·m/s towards the North (c) No matter in which direction our hero decides to throw his book, what is the total momentum of the system of particles consisting of the book and our hero combined as they are slipping across the ice immediately AFTER the thrown physics book leaves our hero's hand? HINT: Use the Principle of Conservation of Momentum. kg·m/s towards the North (d) Our hero manages to throw the book with a speed of 9.88 m/s (this speed is measured by an observer who is stationary on the ice). What will then be the amount of momentum possessed by the book alone immediately AFTER the throw? kg·m/s (e) If our hero has thrown the book towards the North at the speed of 9.88 m/s, what will then be the momentum of our hero immediately after the throw? HINT: From a correct answer to part (c), you know what will be the total momentum of the hero and the book COMBINED immediately after the throw; also, from a correct answer to part (d), you know what will be the amount of momentum of the BOOK ALONE after the throw. If the book's momentum is Northward, "adding the arrows" should easily tell you the momentum of the hero after the throw. kg·m/s

Explanation / Answer

(a)

Throw the physics book towards the North.

(b)

momentum = (mass)(velocity)

p = (60 + 2.4 kg)(0.38 m/s)

p = 23.712 kg-m/s

(c)

Since momentum is conserved, the momentum afterward would be the same as part (b):

23.712 kg-m/s.

(d)

p = (2.4 kg)(9.88 m/s)

p = 23.712 kg-m/s (In general, this would usually not be the same as the momenta in parts (b) and (c) -- the numbers happened to conveniently work out this way for this problem.)

(e)

(total momentum before) = (momentum of thrown book) + (momentum of hapless hero)

23.712 kg-m/s = 23.712 kg-m/s + p

p = 0

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