1. What is the ejection speed of a ball if it was shot horizontally 1 meter abov

Question

1. What is the ejection speed of a ball if it was shot horizontally 1 meter above the floor, and touched it 2.5 meters further from the vertical projection of the ejection point? Use 9.810 m/s^2 for the acceleration of gravity.

2. Identical shot from the apparatus of Problem 1 was caught by the pendulum basket initially at rest. From angular momentum conservation, find the angular speed of the system "pendulum + ball" immediately after the hit. The following necessary data are given: mass of ball 0.060 kg, mass of pendulum 0.200 kg, distance pivot - center of mass of pendulum 0.250 m, distance pivot - center of pendulum basket 0.300 m, period of free swing of pendulum 1.0 s.

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Information Given in Text for Above:

Symbols Defined:

v1= initial speed of the ball

w2=angular speed of pendulum +ball after the collision

m=mass of the ball

M= mass of the pendulum

I= moment of inertia of the pendulum about the pivot

L= distance from pivot to center of basket

R= distance from pivot to center of mass of pendulum

H= rise of the center of the basket until it halts

h= height of gun above the floor

d= horizontal distance free falling ball goes before hitting floor

T= period of the swinging pendulum

Projectile Speed Equation: v1=d[ Sqrt(g/2h)]

Moment of Inertia Equation: I=MgR(T^2)/4pi^2

Other given Equations:

1. Iw2+m(L^2)w2=mLv1

2. .5I(w2^2)+.5m(L^2)(w2^2)-mgL-MgR

3. -mgcosTHETA-MgRcosTHETA

Explanation / Answer

Let teh speed of ejection be v

time taken = t

so 2.5 = v*t

also along y direction

acceleration = 9.8

displacement = 1

1 = 0.5 * 9.81 * t^2

so, t = 0.451524 secs

so ejection sped v = 5.5368 m/sec

2)

intial angular momentum = 0.06 * 5.5368 * 0.3 = 0.0996624

final angular momentum = 0.06 *0.3^2 * w + 0.2*0.25^2 * w

inital momentum = final

so, 0.0179 * w = 0.0996624

w = 0.1796 rad / sec

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