# Collision

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?

________rad/s

(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?

________J

https://brainmass.com/physics/conservation-of-momentum/glancing-collision-two-pucks-255242

## SOLUTION This solution is **FREE** courtesy of BrainMass!

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 puck from the axis

L1 = 0.08 x 1.5 x (0.036 + 0.06 - 0.0375) = 0.007 kgm2/sec

Direction of this angular momentum is clockwise.

Angular momentum of larger puck = 0 (as it is at rest)

Net angular momentum of the system = 0.007 kgm2/sec

b) Moment of inertia of the smaller puck about its own axis = 2/5 m1R12 = 2/5 x 0.08 x 0.0362 = 4.15 x 10-5 kgm2

Applying parallel axis theorem, moment of inertia of the smaller puck about the axis passing through the CM of the system, perpendicular to the plane in which the centers of the pucks lie = MI about its own axis + its mass x (perpendicular distance of its own axis from the new axis)2 = I1 = 4.15 x 10-5+ 0.08x(0.036 + 0.06 - 0.0375)2 = 31.5x10-5 kgm2

Moment of inertia of the larger puck about its own axis = 2/5 m2R22 = 2/5 x 0.125 x 0.062 = 18 x 10-5 kgm2

Applying parallel axis theorem, moment of inertia of the larger puck about the axis passing through the CM of the system, perpendicular to the plane in which the centers of the pucks lie = I2 = 18 x 10-5+ 0.125 x 0.03752 = 35.6x10-5 kgm2

Net moment of inertia = I = I1 + I2 = 31.5x10-5 + 35.6x10-5 = 67.1 x 10-5 kgm2

Net angular momentum after collision = L2 = IÏ‰ where Ï‰ = angular speed immediately after collision

L2 = 67.1 x 10-5 x Ï‰

As the system is isolated rotationally (no external torque), conservation of angular momentum will apply. Hence,

Net angular momentum of the system after collision = Net angular momentum before collision

Or 67.1 x 10-5 x Ï‰ = 0.007

Or Ï‰ = 0.007/67.1 x 10-5 = 10.43 rad/sec

c) Before collision linear momentum of smaller puck = m1v1 = 0.08x1.5 = 0.12 kgm/sec

Before collision linear momentum of smaller puck = m2v2 = 0 (as it is at rest)

Linear momentum of the system after collision = (m1+m2)v = 0.205v where v = velocity of the system after collision

As the system is isolated linearly (no external force), conservation of linear momentum will apply. Hence,

0.205v = 0.12 or v = 0.59 m/s

d) Final KE = Final linear KE + Final rotational KE = Â½ (m1+m2)v2 + Â½ IÏ‰2 = Â½ x 0.205 x 0.592 + Â½ x 67.1 x 10-5 x 10.432 = 0.0357 + 0.0365 = 0.0722 J

e) Before collision KE of the system = Â½ m1v12 = Â½ x 0.08 x 1.52 = 0.09 J

Loss of KE during collision = 0.09 - 0.0722 = 0.0178 J

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