A marble of mass 17.0 g is fired straight upward from a spring-loaded gun. Initi

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. 1 – 2 – (a) At what speed does the marble leave the gun? (b) How high above the ground does the marble go? (Ignore air resistance.)

Explanation / Answer

Here ,

speed of the marble when the leave the gun is v

Using conservation of energy

0.5 * 0.017 * v^2 = 0.50 * 10.2 * (0.252)^2

v = 6.17 m/s

the speed of the marble is 6.17 m/s

b)

height of marble = u^2/(2g) + 0.386

height of marble = 6.17^2/(2 * 9.8) + 0.386

height of marble = 2.33 m

the height of marble is 2.33 m

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