Two skaters stand facing each other. One skater's mass is 60 kg, and the other's
1 answer:
Answer:
v' = 2.4 m/s
Explanation:
Given that,
Mass of one skater, m = 60 kg
Mass of the other's skater, m' = 60 kg
The two skaters push off each other. After the push, the smaller skater has a velocity of 3.0 m/s.
When there is no external force acting on a system, the momentum remains conserved. It means initial momentum is equal to the final momentum. Let v' is the velocity of the larger skater.
mv = m'v'
So, the velocity of the larger skater is 2.4 m/s.
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The sleds speed when the spring returns toits uncompressed length is v = 0.03 m/s.
<u>Explanation</u>:
Given,
force constant = 42 N/cm = 0.42 N/m, mass m = 68 kg, spring x = 0.39 m
The potential energy, U, stored in the spring is
U = 1/2 kx^2
= 1 / 2 0.42
(0.39)^2
= 0.032 J
All its potential energy has been converted into kinetic energy since it has a uncompressed length.
K = 1/2 mv^2
v = sqrt (2K / m)
= √(( 2 0.032) / 68)
v = 0.03 m/s .