Question

A 3.13 kg particle has a velocity v3.13 kg = a ˆi + b ˆj, where a = 24.2 m/s and b = 1.35 m/s, and a 2.81 kg particle has a velocity v2.81 kg = c ˆi + d ˆj, where c = 1.27 m/s and d = 8.54 m/s. Find the speed of the center of mass. Answer in units of m/s.

Answer 1

The speed of the center of mass is
`(20.08\hat i+4.74 \hat j)m/s`

Step-by-step explanation:

The speed of center of mass:

When a system of particles is moving, then the center of mass moves along with the system.

The center of mass velocity is ratio of the sum of momentum ( mass times velocity) to total mass of the system.

`v_(cm)=(v_1m_1+v_2m_2+......+v_nm_n)/(m_1+m_2+......+m_n)`

Given that,

A 3.13 kg particle has a velocity
`a\hat i +b\hat j`, where a= 24.2 m/s and b=1.35 m/s

and other particle has a velocity
`c\hat i+ d\hat j`, where c= 1.27 m/s and d=8.54 m/s

Here,
`v_1 = 24.2 \hat i +1.35 \hat j` ,
`m_1` = 3.13 kg

`v_2 = 1.27 \hat i +8.54\hat j` ,
`m_2` = 2.81 kg

The speed of the center of mass is

`v_(cm)=(v_1m_1+v_2m_2)/(m_1+m_2)`

`=((24.2 \hat i +1.35 \hat j) .3.13+( 1.27 \hat i +8.54\hat j). 2.81)/(3.13+2.82)`

`=((75.746\hat i +4.2255\hat j) +( 3.5687\hat i +23.9974\hat j))/(5.95)`

`=20.08\hat i+4.74 \hat j` m/s