A marble of mass 17.0 g is fired straight upward from a spring-loaded gun. Initially, the marble, resting on top of the spring is 38.6 cm above the ground. The spring is ideal and massless. The force constant of the spring is 10.2 N/m. The spring is initially compressed 25.2 cm. The marble exits the gun just as the spring achieves its unstretched/uncompressed length.
a) At what speed does the marble leave the gun?
b) How high above the ground does the marble go? (Ignore air resistance.)

Question

A marble of mass 17.0 g is fired straight upward from a spring-loaded gun. Initially, the marble, resting on top of the spring is 38.6 cm above the ground. The spring is ideal and massless.  The force constant of the spring is 10.2 N/m. The spring is initially compressed 25.2 cm. The marble exits the gun just as the spring achieves its unstretched/uncompressed length.
a) At what speed does the marble leave the gun?
b) How high above the ground does the marble go?  (Ignore air resistance.)

irishboy123 · Answer

"The spring is initially compressed 25.2 cm"

means what? ie what is height above ground at that time?  need to know compression in spring at that time.

simplest approach:

calc gravitational potential (mgh) of particle and potential spring energy (1/2) k x^2 and particles kinetic energy (1/2) m v^ 2.

do this at the various points described in the question......

anonymous · Answer

I think it means that the spring is compressed by 25.2 cm making the marble rest 13.4 cm above the ground before its launched. into the air a certain distance. But I can't see a way to calculate the height without speed, and with very little understanding of physics i'm not sure how to calculate the speed with the information given.