Question

(6%) Problem 6: The figure shows the view looking down onto an air hockey table. A puck collides with a stationary bar. The collision is completely inelastic. 1 The bar has mass M= 1.33 kilograms and length L = 1.36 meters. The puck has mass m = 0.422 kilograms and is initially moving at vi = 7.38 perpendicular to the bar. The puck collides with the bar at a location of b=0.455 meters from the center of the bar. Note that the size of the puck is negligible compared to the length of the bar. 50% Part (a) Calculate the center of mass speed (in m/s) of the puck-and-bar system just after the collision. Ve=1.78 Correct! >> 50% Part (b) Calculate the angular speed (rad/s) about the center of mass of the puck-and-bar system just after the collision. Grade Summary Deductions 0% Potential 100% Activate Windows sino) cos() tan() * ( 7 8 9 HOME Go tSubmissionsto activate

Answer 1

a) vf = (m*vi + M*0)/(m + M)

= (0.422*7.38 + 0 )/(0.422 + 1.33)

= 1.78 m/s

b) let x is the distance of center of the ball and the rod from the place of the puck.

x = (m*0 + M*b)/(m + M)

= (0 + 1.33*0.455)/(0.422 + 1.33)

= 0.3454 m

moment of inertia of the system about its center of mass,

Icm = M*L^2/12 + M*(b-x)^2 + m*x^2

= 1.33*1.36^2/12 + 1.33*(0.455 - 0.3454)^2 + 0.422*0.3454^2

= 0.2713 kg.m^2

let w is the final angular speed of the puck-and-bar system just after the collision.

Apply conservation of angular momentum

Lf = Li

Icm*w = m*vi*x

w = m*vi*x/Icm

= 0.422*7.38*0.3454/0.2713

= 3.96 rad/s <<<<<<<<<---------------------Answer