A hockey puck (mass mA=0.20kg) move along the ice in the x- direction at speech

Question

A hockey puck (mass mA=0.20kg) move along the ice in the x- direction at speech of 6.0 m/s. it strikes a second puck(mB=0.3kg) which is initially at rest. After the collision it is observed that puck & moves off at an angle of 15 degree above the a-axis while puck B move at the angle of 9 degree below the x-axis. Determine final velocities for the two pucks. Calculate the initial and final translational kinetic energy for the system. If the diameters of each of the pucks Is 6 cm. calculate the final angular velocity for each, assuming there is no initial rotation of the incoming puck.

Explanation / Answer

mA = 0.2 Kg
mB = 0.3 kg
vA = 6.0 m/s

A)
Using Momentum Conservation
In X axis
mA * v = mA * Va*cos(15) + mB * Vb*cos(9)
0.2 * 6.0 = 0.2 *Va*cos(15) + 0.3 * Vb*cos(9) -----------1

In Y axis
mA * Va*sin(15) = mB * Vb*sin(9)
0.2 * Va*sin(15) = 0.3 * Vb*sin(9) ------------2

Using Equation 1 & 2
Va = 2.31 m/s
Vb = 2.55 m/s

Initial Translational Kinetic Energy = 1/2 * mA*v^2
Initial Translational Kinetic Energy = 1/2 * 0.2 * 6^2 J
Initial Translational Kinetic Energy = 3.6 J

Final Translational Kinetic Energy = 1/2 * mA*Va^2 + 1/2 * mB*Vb^2
Final Translational Kinetic Energy = 1/2 * 0.2 * 2.31^2 + 1/2 * 0.3 * 2.55^2
Final Translational Kinetic Energy = 1.51J

(B)
We know,
V = r*W

r = 6/2 = 3 cm
r = 0.03 m

Final Angular Velocity of Pluck A = Va/r
wA = 2.31/0.03 rad/s
wA = 77 rad/s

Final Angular Velocity of Pluck B = Vb/r
wB = 2.55/0.03 rad/s
wB = 85 rad/s

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