Question

A 45.5 kg student runs down the sidewalk and jumps with a horizontal speed of 4.25 m/s onto a stationary skateboard. The student and skateboard move down the sidewalk with a speed of 4.00 m/s. a) Find the mass of the skateboard. Answer in units of kg. 006 (part 2 of 2) 10.0 points b) How fast would the student have to jump to have a final speed of 6.08 m/s

Answer 1

Explanation:

Below is the solution to the question.

Answer 2

The mass of the skateboard and the velocity of the student for a certain case is required.

The mass of the skateboard is 2.84 kg.

The velocity of the student should be 6.46 m/s.

$m_1$ = Mass of student = 45.5 kg

$m_2$ = Mass of skateboard

u = Velocity of student = 4.25 m/s

v = Combined velocity of student and skateboard = 4 m/s

As momentum is conserved we have

$m_1u=m_1v+m_2v\\\Rightarrow m_2=\dfrac{m_1u-m_1v}{v}\\\Rightarrow m_2=\dfrac{45.5\times 4.25-45.5\times 4}{4}\\\Rightarrow m_2=2.84\ \text{kg}$

Now $v=6.08\ \text{m/s}$

$u=\dfrac{(m_1+m_2)v}{m_1}\\\Rightarrow u=\dfrac{(45.5+2.84)\times 6.08}{45.5}\\\Rightarrow u=6.46\ \text{m/s}$

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