A man of mass m 1 5 70.0 kg is skating at v1 5 8.00 m/s behind his wife of mass m 2 5 50.0 kg, who is skating at v2 5 4.00 m/s. Instead of passing her, he inadvertently collides with her. He grabs her around the waist, and they maintain their balance. (a) Sketch the problem with before - and - after diagrams, representing the skaters as blocks. (b) Is the collision best described as elastic, inelastic, or perfectly inelastic? Why? (c) Write the general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf . (d) Solve the momentum equation for vf . (e) Substitute values, obtaining the numerical value for vf , their speed after the collision.

A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)

B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist

C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf

Say mass of husband is m1

Mass of the wife is m2

Velocity of the husband is v1

Velocity of the wife is v2

According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf

The momentum equation is

m1v1+m2v2= (m1+m2)vf

D. To solve for vf we need to make it subject of formula

vf= {(m1v1) +(m2v2)}/(m1+m2)

E. Substituting our given data

vf=

{(1570*58)+(2550*54)}/(1570+2558)

vf=91060+137700/4120

vf=228760/4120

vf=55.52m/s

Their speed after collision is 55.52m/s

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid