Question

An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into a 10.0-kg pendulum bob that is attached to an arm of length L. Assume that the 9-mm bullet has a mass of 6.00 g and the 0.44-caliber bullet has a mass of 12.0 g . If the 9-mm bullet causes the pendulum to swing to a maximum angular displacement of 4.30∘ and the 0.44-caliber bullet causes a maximum displacement of 10.1∘, find the ratio of the initial speed of the 9-mm bullet to the speed of the 0.44-caliber bullet, (v0)9/(v0)0.44.

Answer 1

Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

$v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}$

The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.

h = L * (1 - cos(a))

$v2 = \sqrt{\frac{g * L * (1 - cos(a)) * (M+ mb)^2}{\frac{1}{2} * mb^2}}$

For the 9 mm:

$v2 = \sqrt{\frac{9.81 * L * (1 - cos(4.3)) * (10+ 0.006)^2}{\frac{1}{2} * 0.006^2}} = \sqrt{L} * 391$

For the 0.44 caliber:

$v2 = \sqrt{\frac{9.81 * L * (1 - cos(10.1)) * (10+ 0.012)^2}{\frac{1}{2} * 0.012^2}} = \sqrt{L} * 460$

The ratio is 460 / 391 = 1.176