Using a different launcher in the Ballistic Pendulum Experiment a ball of mass 5

Question

Using a different launcher in the Ballistic Pendulum Experiment a ball of mass 5.13 grams is fired into the "bob" of the pendulum. The pendulum deflects 9.5'. The distance from the axis of the axis of rotation to the mass of the pendulum is 0.265 meters and the mass of the pendulum is 240.3g. The distance from the axis to the center of mass of the ball is .286 meters. The period of oscillation of the pendulum and embedded ball is 1.13 seconds. What is the muzzle velocity of the ball the instant before it hits the pendulum "bob"?

Explanation / Answer

m v = (m + M) V conservation of momentum V = m v / (m + M) speed of pendulum after collision 1/2 (M + m) V^2 = (m + M) g h conservation of energy after collision V^2 = 2 g h h is height to which pendulum rises V^2 = m^2 v^2 / (m + M)^2 v^2 = (m + M)^2 * 2 g h / m^2 v = (m + M) / m * (2 g h)^1/2 initial speed of ball h = L (1 - cos theta) substitute in preceding equation If we use the equation for a simple pendulum to get the distance L from the axis of rotation to the center of mass then T = 2 pi (L / g)^1/2 L = T^2 g / (4 * pi^2) = .317 m Apparently the distance to the C.M. is not the distance to the ball

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