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

Answer:

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

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

$m_{b}v_{b}\cos\theta= m_{c}v_{c}$

$v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}$

Put the value into the formula

$v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}$

$v_{c}=0.0345\ m/s$

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