Answer:
Final speed, v = 1.25 m/s
Explanation:
Given that,
Mass of skater A, [tex]m_A=32.7\ kg[/tex]
Initial speed of skater A, [tex]u_A=2.3i\ m/s[/tex] (x axis)
Mass of skater B, [tex]m_B=93.3\ kg[/tex]
Initial speed of skater B, [tex]u_B=1.51j\ m/s[/tex] (y axis)
It is mentioned that the two skaters collide and cling together. It is case of inelastic collision in which momentum remains conserved. Let V is the final speed of the couple. It is given by :
[tex]m_Au_A+m_Bu_B=(m_A+m_B)V[/tex]
[tex]32.7\times 2.3i+93.3\times 1.51j=(32.7+93.3)V[/tex]
[tex]V=\dfrac{75.21i+140.88j}{126}[/tex]
[tex]V=(0.59i+1.11j)\ m/s[/tex]
[tex]|V|=1.25\ m/s[/tex]
So, the final speed of the couple is 1.25 m/s. Hence, this is the required solution.
