Question

Two persons on ice skates stand face to face and then push each other away (see figure below). Their masses are 80 and 95 kg.

Find the ratio of their speeds immediately afterward.
v80 kg
v95 kg
=


Which person has the higher speed?
80-kg skater
95-kg skater

Answer 1

This question involves the concepts of the law of conservation of momentum.

The ratio of their speeds will be "-1.19".

"80-kg skater" has the higher speed.

LAW OF CONSERVATION OF MOMENTUM:

According to the Law of Conservation of Momentum, the total momentum of an isolated system always remains constant. Therefore,

$Total\ Momentum\ Before\ Collision =Total\ Momentum\ After\ Collision \\\\m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

• m₁ = mass of skater 1 = 80 kg
• m₂ = mass of skater 2 = 95 kg
• u₁ = speed of skater 1 before collision = 0 m/s
• u₂ = speed of skater 2 before collision = 0 m/s
• v₁ = v₈₀ = speed of skater 1 after collision
• v₂ = v₉₅ = speed of skater 2 collision

Therefore,

$(80\ kg)(0\ m/s)+(95\ kg)(0\ m/s)=(80\ kg)(v_{80})+(95\ kg\ kg)(v_{95})\\\\\frac{v_{80}}{v_{95}}=-\frac{95\ kg}{80\ kg\ kg}\\\\\frac{v_{80}}{v_{95}}=-1.19$

It is clear that the speed of skater A (80 kg skater) will be higher.

Learn more about the Law of Conservation of Momentum here:

https://brainly.com/question/1113396