Answer: [tex]= (1.31i+2.42j)m/s\\[/tex]
Explanation:
GIVEN DATA:
[tex]m1 = 2.03kg\\m2 = 3.02kg\\v1 = 1.90 i - 2.91 j) m/s\\v2 = (0.92 i -6.00 j) m/s\\[/tex]
Solution:
the velocity for centre of the mass
[tex]vcm= \frac{m1v1+m2v2}{m1+m2}[/tex]
input the values into the formula.
[tex]vcm=\frac{(2.03)(1.90 i - 2.91 j)+(3.02)(0.92 i -6.00 j) }{2.03+3.02}[/tex]
open the bracket
[tex]= \frac{3.86i-5.91j +2.78i+18.12j}{5.05}[/tex]
collect like terms
[tex]= \frac{6.64i+12.21j}{5.05}[/tex]
[tex]= (1.31i+2.42j)m/s[/tex]
total momentum
[tex]p= (m1+m2)vcm\\= (2.03+3.02) (1.31i+2.42j)\\= 5.05 (1.31i+2.42j)\\=(6.62i+12.22j)kg.m/s[/tex]
