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s344n2d4d5 [400]

3 years ago

9

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 c

omponent of the velocity of the center of mass.

1 answer:

hammer [34]3 years ago

4 0

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

Explanation:

GIVEN DATA:

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

<u><em>Solution:</em></u>

the velocity for centre of the mass

$vcm= \frac{m1v1+m2v2}{m1+m2}$

input the values into the formula.

$vcm=\frac{(2.03)(1.90 i - 2.91 j)+(3.02)(0.92 i -6.00 j) }{2.03+3.02}$

open the bracket

$= \frac{3.86i-5.91j +2.78i+18.12j}{5.05}$

collect like terms

$= \frac{6.64i+12.21j}{5.05}$

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

total momentum

$p= (m1+m2)vcm\\= (2.03+3.02) (1.31i+2.42j)\\= 5.05 (1.31i+2.42j)\\=(6.62i+12.22j)kg.m/s$

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