Part-A : If the rod makes one revolution every 0.736 s after the puck is caught, then the puck's speed just before it hit the rod which is given as -
f = (1 x 2) / (0.736 s) = 8.53 rad/s
using conservation of momentum, we have
mr v0,r + mp v0,p = (mr + mp) vf
where, v0,r = initial speed of rod = 0 m/s
we know that, angular velocity is given by -
f = vf / r (8.53 rad/s) = vf / (2 m)
vf = 17.06 m/s
then, we get
(1.3 kg) (0 m/s) + (0.163 kg) v0,p = [(1.3 kg) + (0.163 kg)] (17.06 m/s)
(0.163 kg) v0,p = (24.9 kg.m/s)
v0,p = 152.7 m/s
A local ice hockey team has asked you to design an apparatus for measuring the speed...
An 160.0 g hockey puck slides along an essentially frictionless ice rink with speed 4.70 m/s. A hockey player uses her stick to do –1.20 J of work on the puck. What is the puck's speed after she has done this work? A. 0 m/s B. 2.66 m/s C. 3.50 m/s D. 4.53 m/s