p08_027.pdf

Fromtheseweﬁnd x = v 0 2 h/g . We note from this that the distance to the landing point is directly

proportional to the initial speed. We denote v 01 be the initial speed of the ﬁrst shot and x 1 =1 . 93 m

be the horizontal distance to its landing point; similarly, v 02 is the initial speed of the second shot and

x 2 =2 . 20 m is the horizontal distance to its landing spot. Then

v 02

v 01 = x 2

⇒

v 02 = x 2

x 1

When the spring is compressed an amount , the elastic potential energy is 2 k 2 . When the marble

leaves the spring its kinetic energy is 2 mv 0 . Mechanical energy is conserved: 2 mv 0 = 2 k 2 ,andwesee

that the initial speed of the marble is directly proportional to the original compression of the spring. If

1 is the compression for the ﬁrst shot and 2 is the compression for the second, then v 02 =( 2 / 1 ) v 01 .

Relating this to the previous result, we obtain

x 1 1 = 2 . 20m

1 . 93m

(1 . 10cm) = 1 . 25 cm .

27. The distance the marble travels is determined by its initial speed (and the methods of Chapter 4), and

the initial speed is determined (using energy conservation) by the original compression of the spring. We

denote h as the height of the table, and x as the horizontal distance to the point where the marble lands.

Then x = v 0 t and h = 2 gt 2 (sinc e the vertical component of the marble’s “launch velocity” is zero).

x 1 v 01 .

2 = x 2