40 kg girl skates at 3.5 m/s on ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold each other, what is the speed of the couple?

We need to consider a direct central impact, as the couple remains together after the impact there is no restitution. Therefore impact the perfectly plastic.

Therefore the initial data:

v1 = velocity of the 40 [kg] girl = 3.5[m/s]

m1 = mass of the first girl = 40 [kg]

m2 = mass of the second girl = 65 [kg]

v2 = velocity of the two girls after the collision.

[tex]m_{1}*v_{1}+m_{2}*v_{2}=(m_{1}+m_{2})*v_{2}\\(40*3.5)+(65*0)=(65+40)*v_{2}\\v_{2}=140/105\\v_{2}=1.33[m/s][/tex]

asuwafohjames asuwafohjames · Answer 2 · 2022-01-28T12:07:05+01:00

The speed of the couple is 1.33 m/s

To calculate the speed of the couple, we use the formula below.

Formula:

mu+m'u' = V(m+m')............... Equation 1

Where:

m = mass of the girl
m' = mass of the girl's friend
u = initial velocity of the girl
u' = initial velocity of the girl's friend
V = speed of the couple.

make V the subject of the equation

V = (mu+m'u')/(m+m')................... Equation 2

From the question,

Given:

m = 40 kg
m' = 65 kg
u = 3.5 m/s
u' = 0 m/s (standing still)

Substitute these values into equation 2.

V = [(40×3.5)+(65×0)]/(40+65)
V = 140/105
V = 1.33 m/s

Hence, The speed of the couple is 1.33 m/s.

Learn more about speed here: https://brainly.com/question/4931057