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A puck of mass 80.0 g and radius 3.60 cm slides along an air table at a speed of 1.50 m/s as shown in Figure P11.34a. It makes a glancing collision with a second puck of radius 6.00 cm and mass 125.0 g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and spin after the collision (Fig. P11.34b).

Figure P11.34
(a) What is the angular momentum of the system relative to the center of mass?
________kg m2/s
(b) What is the angular speed about the center of mass after collision?
(c) What is the velocity of the center of mass?
________ m/s
(d) What is the final kinetic energy?
________ J
(e) How much energy is lost in the collision?

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Solution Summary

The angular momentum and speed of a collision are determined. A detailed step by step solution with a figure provided.

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Let us first determine the centre of mass of the two pucks. We take the line passing through the larger puck, parallel to the direction of motion of the smaller puck as the reference.

Let the distances of the centers of the two pucks from the reference line be d1 and d2 and the distance of the centre of mass of the system of two pucks from the reference line at the instant of the rim collision be d. Then, by definition:

d = (m1d1+m2d2)/(m1+m2) = (0.08x0.096 + 0.0125x0)/(0.08+0.125) = 0.00768/0.205 = 0.0375 m or 3.75 cm

a) Angular momentum of smaller puck about an axis passing through the CM perpendicular to the plane in which the centers of the pucks lie, just before the collision at the rims = L1 = m1vr1 where r1 = distance of the centre of the ...

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