Question

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.5 kg to a friend standing in front of him. The student and the skateboard have a combined mass of mc = 105 kg and the book leaves his hand at a velocity of vb = 2.97 m/s at an angle of 26° with respect to the horizontal.
(a)Write an expression for the magnitude of the velocity of the student,vs, after throwing the book(b)Calculate the magnitude of the velocity of the student,vs, in meters per second?(c)What is the magnitude of the momentum,pe, which was transferred from the skateboard to the the Earth during the time the book isbeing thrown (in kilogram meters per second)?

Answer 1

Answer:

a) $v_s=\frac{m_bv_bcos\theta}{m_c}$

b)  $v_s=0.0385$ m/s

c) $P_e$= 1.952 kg.m/s

Explanation:

Given:

Mass of the skateboard, $m_b$ = 1.5 kg

Combined mass, $m_c$ = 105 kg

velocity of the book, $v_b$ = 2.97 m/s

angle, θ = 26°

thus,

horizontal component of the velocity = vcosθ

vertical component of velocity = vsinθ

a) Applying the concept of conservation of momentum

$m_bv_bcos\theta=m_cv_s$

where,

$v_s$ is the velocity of the the student

thus,

$v_s=\frac{m_bv_bcos\theta}{m_c}$      ............(1)

b) on substituting the values, in the equation we get the magnitude

$v_s=\frac{1.5\times2.97cos26^o}{105}$

or

$v_s=0.0385$ m/s

c) Now, the momentum (P) transferred is given as:

P = mass × velocity

on substituting the values, we get

$P_e$= mass × velocity

or

$P_e$= $m_b\times v_b\times \sin\theta$

on substituting the values, we get

$P_e$= $1.5\times 2.97\times \sin26^o$

or

$P_e$= 1.952 kg.m/s