Answer:
2.39 kg
Explanation:
There is conservation of momentum here in this problem so we will use the following problem:
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
where the mass of the student [tex]m_1[/tex] is 48.5 kg,
the mass of the skateboard [tex]m_2[/tex] is [tex]m_2[/tex] kg,
the initial speed of the student [tex]u_1[/tex] is 4.25 m/s; and
the speed of the student and skateboard [tex]v[/tex] is 4.05 m/s.
So substituting the given values in the above formula to get:
[tex](48.5*4.25) + (m_2 * 0) = (48.5 + m_2 ) * 4.05[/tex]
[tex]206.125=196.425+4.05m_2[/tex]
[tex]206.125 - 196.425 = 4.05m_2[/tex]
[tex]m_2=\frac{9.7}{4.05}[/tex]
[tex]m_2=2.39[/tex]
Therefore, the mass of the skateboard is 2.39 kg.
