Question

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him. The student and the skateboard have a combined mass of mc = 112 kg and the book leaves his hand at a velocity of vb = 3.61 m/s at an angle of 31° with respect to the horizontal.

Randomized Variables :

mt = 1.05 kg

mc = 104 kg

Vb = 2.25 m/s

θ = 22 degrees


What is an expression for the magnitude of the velocity the student has after throwing the book?

Answer 1

The velocity of the student has after throwing the book is 0.0345 m/s.

Step-by-step explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal direction

`m_(b)v_(b)\cos\theta= m_(c)v_(c)`

`v_(s)=(m_(b)v_(b)\cos\theta)/(m_(c))`

Put the value into the formula

`v_(c)=(1.25*3.61*\cos31)/(112)`

`v_(c)=0.0345\ m/s`

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.