A uniform stick 1.3m long with a total mass of 270g is pivoted at its center. A

Question

A uniform stick 1.3m long with a total mass of 270g is pivoted at its center. A 3.0g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250m/s and leaves at 140m/s. (Figure 1)

With what angular speed is the stick spinning after the collision?

A uniform stick 1.3m long with a total mass of 270g is pivoted at its center. A 3.0g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250m/s and leaves at 140m/s. (Figure 1) With what angular speed is the stick spinning after the collision?

Explanation / Answer

Angular speed = v/r, where r is the distance between the pivot and the point of impact.

If momentum is conserved, m1v1 = m1v2 + m2vstick

(3)(250) = (3)(140) + (270)vstick

750 = 420 + 270vstick

330 = 270vstick

1.22 m/s = vstick

= vstick/(.325 m) = 1.22/.325 = 3.75 s-1

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