Answer:
a). 6.09 m/s
b). α=54.35
Explanation:
The collision is an elastic so both skaters return their mass but change their velocities
[tex]m_{d}*v_{id}+m_{r}*v_{ir}=m_{d}*v_{fd}+m_{r}*v_{fr}[/tex]
The initial velocity of Daniel is zero because the before the collision he is in rest
[tex]70kg*0+45kg*12\frac{m}{s}=70kg*v_{fdx}+45kg*7\frac{m}{s}*cos(52.1)[/tex]
To determine the final velocity of Daniel
[tex]540=70*v_{fdx}*cos(\alpha)+193.49[/tex]
[tex]v_{fdx}*cos(\alpha)=4.95[/tex]
So in the axis 'y'
[tex]70*v_{fdy}*sin(\alpha)=45*7*sin(52.1)[/tex]
[tex]v_{fdy}*sin(\alpha)=3.55[/tex]
a).
The magnitude of the velocity Daniels after the collision is
[tex]V=\sqrt{v_{fdx}^2+v_{fdy}^2}[/tex]
[tex]V=\sqrt{4.95^2+3.55^2}[/tex]
[tex]v=6.09\frac{m}{s}[/tex]
b).
The direction of the velocity after the collision is
[tex]=\frac{cos(\alpha)}{sin(\alpha)}[/tex]
[tex]tan\alpha =\frac{v_{dfx}}{v_{dfy}}[/tex]
[tex]tan(\alpha )=\frac{4.95}{3.55}[/tex]
[tex]\alpha =tan^-1*(1.39)[/tex]
[tex]\alpha =54.35[/tex]
