Answer:
6.18 m/s
Explanation:
Roller skate collision
The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;
x-axis component form (+x east);
[tex]P_{Miy}[/tex] + [tex]p_{Piy}[/tex] + [tex]j_{y}[/tex]= [tex]P_{Mfy}[/tex] +[tex]P_{pfy}[/tex]
[tex]m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin[/tex]Ф
60 ·8 + 0 = (60 + 80)[tex]V_{f}sin[/tex]Ф
480 = 140[tex]V_{f} sin[/tex]Ф................. (I)
y-axis component form (+y north);
[tex]P_{Mix}[/tex] + [tex]p_{Pix}[/tex] + [tex]j_{x}[/tex] = [tex]P_{Mfx}[/tex]+ [tex]P_{pfx}[/tex]
[tex]m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos[/tex]Ф
0 + 80.9 = (60 + 80)[tex]V_{f}cos[/tex]Ф
720= 140[tex]V_{f}cos[/tex]Ф
140Vf=[tex]\frac{720}{cos}[/tex]Ф......................................(2)
Substituting (2) into (1) to give the angle;
480 = 720tan Ф
Ф = arctan(0.67) =33.69°.......................(3)
Evaluating (1) with (3) gives the velocity magnitude
480 = 140Vfsin 33.69°
Vf=6.18 m/s
note 1:
This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.