Jane and Michael are inventing a game using a spring and marble of mass 200.0 g

Question

Jane and Michael are inventing a game using a spring and

marble of mass 200.0 g in which they plan on employing

inclined planes. As they carefully plan out the design, they

decide to launch the marble along an incline of ?=15o

using the spring of spring constant k=2.10 N/cm as

shown in the figure. The plan is to compress the spring by

7.00 cm and launch the marble along the incline. In order

to complete the design they need to know the speed of

the marble just as it leaves the spring and the final distance the marble will travel up the

incline.

a) Find an expression for the speed of the marble just as it leaves the spring. Hint: Do not

forget about the work done by the gravitational force and assume that the marble slides

without friction.

b) Calculate a numerical value for the speed of the marble just as it leaves the spring.

c) Find the distance up the incline that the marble will travel before coming to rest

momentarily.

Jane and Michael are inventing a game using a spring and marble of mass 200.0 g in which they plan on employing inclined planes. As they carefully plan out the design, they decide to launch the marble along an incline of ?=15o using the spring of spring constant k=2.10 N/cm as shown in the figure. The plan is to compress the spring by 7.00 cm and launch the marble along the incline. In order to complete the design they need to know the speed of the marble just as it leaves the spring and the final distance the marble will travel up the incline. Find an expression for the speed of the marble just as it leaves the spring. Hint: Do not forget about the work done by the gravitational force and assume that the marble slides without friction. Calculate a numerical value for the speed of the marble just as it leaves the spring. Find the distance up the incline that the marble will travel before coming to rest momentarily.

Explanation / Answer

a)
1/2mv^2=1/2kd^2-mgdsin15
d=7 cm
v=(d^2h/m-2dsin15)^1/2

b)
v=0.07(210/0.2-2*0.07*sin15)^1/2
v=2.26 m/s

c)
1/2mv^2=mgh
h=v^2/(2g)
h=5.11/(2*9.8)=0.26 m

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