Question

A 2.03 kg particle has a velocity (1.90 i - 2.91 j) m/s, and a 3.02 kg particle has a velocity (0.92 i 6.00 j) m/s. Find the x component of the velocity of the center of mass.

Answer 1

Answer:
`= (1.31i+2.42j)m/s\\`

Step-by-step explanation:

GIVEN DATA:

`m1 = 2.03kg\\m2 = 3.02kg\\v1 = 1.90 i - 2.91 j) m/s\\v2 = (0.92 i -6.00 j) m/s\\`

Solution:

the velocity for centre of the mass

`vcm= (m1v1+m2v2)/(m1+m2)`

input the values into the formula.

`vcm=((2.03)(1.90 i - 2.91 j)+(3.02)(0.92 i -6.00 j) )/(2.03+3.02)`

open the bracket

`= (3.86i-5.91j +2.78i+18.12j)/(5.05)`

collect like terms

`= (6.64i+12.21j)/(5.05)`

`= (1.31i+2.42j)m/s`

total momentum

`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`