Initial KE = (1/2) m v^2 = 2.5*49 = 122.5 Joules
It lost 122.5 Joules coming to rest. Presumably that was stored as potential energy in the spring.
(1/2) k x^2 = 122.5
.5 k (.05)^2 = 122.5
k = 98,000 Newtons/meter
.9 * 122.5 = 110.25
so
.5 * 5 * v^2 = 110.25
v = 6.64 m/s
A 5kg mass is sliding at a speed of 7 m/s on horizontal sheet of frictionless ice when it smacks in to a wall of the ice rink. There are bumpers made of elastic springs placed along the wall for protection. The mass hits the bumper, compressing it by 5cm and momentarily comes to rest before bouncing off
a) What is the force constant of the spring bumper?
b) If 10% of the mechanical energy is lost during the encounter with the bumper, what is the rebound speed of the mass
2 answers
Thank you
1 answer