Question

The mass of a particular skateboard plus its rider is ms=79.1kg, and the skateboarder is holding a book with mass mb=2.22kg. Initially at rest, the skateboarder tosses the textbook with a velocity of vb=4.44m/s at an angle θ=26.8 above the horizontal, and the book is caught by a friend.

a)Write an expression for the magnitude of the velocity of the student, vs, after throwing the book.

b)Calculate the magnitude, in meters per second, of the velocity of the student.
c)What is the magnitude, in newton seconds, of the momentum, pE, which was transferred to the the Earth during the time the book was being thrown?

Answer 1

The velocity of the student after throwing the book is determined by conservation of momentum. The student's velocity can be calculated by the expression vs = - (mb * vb * cos(θ)) / ms. The momentum transferred to the Earth is equal to the book's momentum after being thrown.

Step-by-step explanation:

The problem relates to conservation of momentum and can be solved using physics concepts, particularly Newton's third law of motion and conservation of linear momentum.

Part a)

To write an expression for the magnitude of the velocity of the student (vs) after throwing the book, you can use conservation of momentum. The initial total momentum of the system is zero since both the skateboarder and the book are at rest. After throwing the book, the total momentum must still be zero. Hence,

Momentum of skateboarder and book before throwing = Momentum of skateboarder after throwing + Momentum of book after throwing

(ms + mb) * 0 = ms * vs + mb * vb * cos(θ)

Therefore, the expression for the skateboarder's velocity (vs) is:

vs = - (mb * vb * cos(θ)) / ms

Part b)

Substituting the given values into the expression,

vs = - ((2.22 kg * 4.44 m/s * cos(26.8°)) / 79.1 kg)

Calculate the magnitude of vs to find the skateboarder's velocity after throwing the book

Part c)

The momentum transferred to the Earth (pE) during the book throw is equal in magnitude to the momentum given to the book since the Earth-skater-book system is closed and we ignore external forces like air resistance. As before, the momentum of the skateboarder and book system before and after the throw must balance each other. Therefore, the magnitude of momentum transferred (pE) is the same as the momentum the book gained,

pE = mb * vb * cos(θ)

Substitute the given values to find pE.