A mass m = 0.5 kg slides in the +x direction on a horizontal frictionless surface with speed vx = 2.3 m/s. It runs into a relaxed spring oriented along the x-axis. This spring had previously been observed to stretch by 0.8 cm when it was oriented vertically with the mass suspended from it.

(a) What is the maximum compression of the spring when the mass runs into it?
How do i find what the constant is because for the equation i got .5mv^2=.5kx^2. Is that right?

( The mass rebounds as a result of having compressed the spring. What is its velocity when it leaves contact with the spring?
Magnitude

m = .5 kg , v = 2.3 m/s

so KE = 1/2 m v^2 = .25 (2.3)^2 = 1.3225

and since after compression KE = 0 and PE spring = 1/2 k x^2

we have

1.3225 = .5 k x^2

or

2.645 = k x^2 so x^2 = 2.645 / k but we need k to find x now.


but you also konw that:




and using Hooke's law u know you have

F = kx so F = mg

so


mg = k x so (convert x=.8cm to .008 meters) k = mg / x = .5 *9.8 / .008= 612.5 = k


so now using the previous equation we have:

x^2 = 2.645 / k = 2.645 / 612.5 = .004318

so x = radical (.004318) = .0657 m = 6.57 cm = x

the second part asks for the speed since there is no friction there is no loss of kinetic energy and the spring merely borrows the kinetic energy and turns it into the potential energy and then releases it so in that sense, the velocity remains the same 2.3 m/s.