A 72.9 kg man is standing on a frictionless ice surface when he throws a 1.6 kg book at 11 m/s.

To answer this question, we can use the principle of conservation of momentum. The initial momentum of the man before throwing the book should be equal to the combined momentum of the man and the book after the throw.

The momentum of an object is calculated by multiplying its mass by its velocity.

The initial momentum of the man can be calculated as:

Initial momentum of the man = mass of the man × velocity of the man

Given:
Mass of the man (m1) = 72.9 kg
Velocity of the man (v1) = 0 m/s (since he is standing still)

So, initial momentum of the man (p1) = m1 × v1 = 72.9 kg × 0 m/s = 0 kg·m/s

Now, let's calculate the momentum of the book after the throw. Since the man throws the book, the momentum of the book will be equal to the momentum of the combined system of the man and the book.

The momentum of the book is given by:

Momentum of the book = mass of the book × velocity of the book

Given:
Mass of the book (m2) = 1.6 kg
Velocity of the book (v2) = 11 m/s

So, momentum of the book (p2) = m2 × v2 = 1.6 kg × 11 m/s

To find the final momentum of the combined system:
Final momentum of the combined system (pf) = initial momentum of the man (p1) + momentum of the book (p2)

pf = p1 + p2

Therefore, final momentum of the combined system (pf) = 0 kg·m/s + (1.6 kg × 11 m/s)

Now, you can calculate the final momentum of the combined system.

Thanks. Did you have a question? I suspect it has to do with conservation of momentum. I will be happy to critique your work. Why would he have a book on an ice surface? Surely not reading it.