In the figure here, two children are playing a game in which they try to hit a s

Question

In the figure here, two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance D = 2.17 m from the edge of the table. Bobby compresses the spring 1.25 cm, but the center of the marble falls 29.2 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring nor the ball encounters friction in the gun.

Explanation / Answer

This system is an isolated system because the only forces that act on the marble are its weight mg and the force of the spring. The shooting speed v0 of the marble can be calculated by using the conservation of energy principle

½m v02 = ½k d2, ==> v0 = d (k/m)½,

where k is the spring constant. The fall of the marble will take a time t, which can be obtained from

h=½ g t2 ==> t = (2h/g)½,

where h is the height of the table. Since the horizontal component of the velocity is constant, range of the marble is

R = v0t = d (k/m)½ (2h/g)½ = d (2kh/gm)½.

Thus, we can write

l - a = d1 (2kh/gm)½ ,

l = d2 (2kh/gm)½ .

Dividing these equations side by side gives

(l-a)/l = d1/d2 ==> d2 = d1 l / (l-a).

Substituting the numerical values, we obtain

d2 = (1.25 cm) (2.17 m)/(2.17 m - 0.292 m) = 1.44 cm.

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