A small metal ball with a mass of m = 53.4 g is attached to a string of length l

Question

A small metal ball with a mass of m = 53.4 g is attached to a string of length l = 1.62 m. It is held at an angle of ? = 73.9° with respect to the vertical.

The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.17 m, and then later it hits the ground. At what distance x will the ball land?

Chrome File Edit View History Bookmarks People Window Help LON-CAPA C ollision pendCChegg.com You https://s2.lite.msu.edu/res/msum agytibo/Mechanics/ParticleSystems/002a.problem#02 ABP Main Menu Contents Grades » Homework set #4 (due 2/ 16 at 11:59PM) » Collision pendulum -C) Timer Notes à Evaluate ,Feedback-Print e,Info Course Contents » Print Info A small metal ball with a mass of m = 53.4 g is attached to a string of length-1.62 m. It is held at an angle of = 73.9° with respect to the vertical Im Im The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h3.17 m, and then later it hits the ground. At what distance x will the ball land? Submit Answer Incorrect. Tries 2/20 Previous Tries Post Discussion Send Feedback

Explanation / Answer

Use the principle of energy conservation to find the kinetic energy of the second ball, use this to find the starting velocity, and then solve the projectile motion problem.

PART A: Potential gravitational energy of first ball

The change height can be found using simple trigonometry.

a is the vertical distance from the ball to the pivot. Change in height (h) is 1.62 - a.
cos() = adj / hyp
cos(73.9) = x / 1.62
a = 0.449 m
h = 1.171 m

Ep = mgh
Ep = 9.8 * 0.0534 * 1.171
Ep = 0.6128 J

PART B: Find initial kinetic energy of second ball

Because of energy conservation, all the gravitational potential energy of the first ball is going to be converted to kinetic energy at the bottom of the swing. If the collision is purely elastic, then all the energy will be transferred from the first ball to the second ball.
Therefore, Ek = Ep = 0.6128 J

PART C: Find initial velocity of second ball

Ek = (1/2)mv²
v = (2 * 0.6128 / 0.0534)
v = 4.79 m/s

PART D: Find how long it will fly before it hits the ground

Ignore the horizontal velocity and look only at vertical velocity: starting vertical velocity is 0m/s, acceleration is 9.8m/s, vertical displacement is 3.17 m, and we are trying to find t.

d = vi * t + 1/2 * a * t²
3.17 = 0 * t + 1/2 * 9.8 * t²
t = (2 * 3.17 / 9.8)
t = 0.804 s

PART E: Find horizontal displacement (assuming the horizontal velocity is constant)

v = d/t
d = 4.79 * 0.804
d = 3.85 m

So the ball will travel 3.850m before landing.

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