Respuesta :
Answer:
The velocity of the student and skateboard together [tex]=1.82\ ms^{-1}[/tex]
Explanation:
Given:
Mass of student [tex]m_1=71\ kg[/tex]
Mass of skateboard [tex]m_2=2.8\ kg[/tex]
Distance between student and skateboard [tex]d=2.75\ m[/tex]
Acceleration of student [tex]a=0.65\ ms^{-2}[/tex]
Finding velocity [tex]v_1[/tex] of the student  before jumping on skateboard
Using equation of motion
[tex]v_1^2=v_0^2+2ad[/tex]
here [tex]v_0[/tex] represents the initial velocity of the student which is [tex]=0[/tex] as he starts from rest.
So,
[tex]v_1^2=0^2+2(0.65)(2.75)[/tex]
[tex]v_1^2=3.575[/tex]
Taking square root both sides:
[tex]\sqrt{v_1^2}=\sqrt[1.7875}[/tex]
∴ [tex]v_1=1.89[/tex]
Finding velocity [tex]v[/tex] of student and skateboard.
Using law of conservation of momentum.
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
Where [tex]v_2[/tex] is initial velocity of skateboard which is [tex]=0[/tex] as it is at rest.
Plugging in values.
[tex]71(1.89)+(2.8)(0)=(71+2.8)\ v[/tex]
[tex]134.19=73.8\ v[/tex]
Dividing both sides by [tex]73.8[/tex]
[tex]\frac{134.19}{73.8}=\frac{73.8\ v}{73.8}[/tex]
∴ [tex]v=1.82[/tex]
The velocity of the student and skateboard together [tex]=1.82\ ms^{-1}[/tex]
