Two children are playing a game in which they attempt to hit a small box on the

Question

Two children are playing a game in which they attempt to hit a small box on the floor using a spring-loaded marble gun fastened on the edge of a frictionless, horizontal table. The box is d = 1.37 m from the base of the table, and the marble is fired horizontally. The first child compresses the spring distance x1 = 3.1 cm, and the marble falls short by ?d1 = 26.7 cm. Find x2, the distance the second child should compress the spring so the same marble falls into the box.

-?cm

HINT: answer is not 0.01215

Explanation / Answer

First, I will assume that the box is on the floor and the height of the top of the table to the center of the box is h. Also, I will assume that there is no resistance in the system. So, when the marble rolls to the edge of the table, it has only horizontal speed. After the lip of the edge it begins to fall. The speed of the marble is related to the compression distance of the spring through conservation of energy and Hooke's Law: .5*m*v^2=.5*k*x^2 at the edge of the table the marble will fly the distance d=v*t and fall the height, h h=.5*g*t^2 What we know about Bobby's attempt: 1.55 - .280=v1*t1 t1=(1.55-.280)/v1 h=.5*9.81*((1.55-.280)/v1)^2 and .5*m*v1^2=.5*k*x1^2 m*v1^2=k*.011^2 the second attempt to score the direct hit: 1.55=v2*t2 t2=1.55/v2 m*v2^2=k*x2^2 h=.5*9.81*(1.55/v2)^2 since the h is constant ((1.55-.280)/v1)^2=(1.55/v2)^2 doing some algebra: v1^2/v2^2=(1.55-.280)/1.55 flipping v2^2/v1^2=1.55/(1.55-.280) dividing the Hooke's Law equations: v1^2/v2^2=0.011^2/x2^2 so x2^2/0.011^2=v2^2/v1^2 x2=sqrt((0.011^2)* 1.55/(1.55-.280)) x2=0.01215m or 1.215 cm

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