Question

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00-m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 1.30-kg rod has a light basket at the other end to catch the 0.163-kg puck. The puck slides across the ice with velocity v vector (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

Answer 1

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 -

$\omega$ _f = (1 x 2 $\pi$ ) / (0.736 s) = 8.53 rad/s

using conservation of momentum, we have

m_r v_0,r + m_p v_0,p = (m_r + m_p) v_f

where, v_0,r = initial speed of rod = 0 m/s

we know that, angular velocity is given by -

$\omega$ _f = v_f / r $\Leftrightarrow$ (8.53 rad/s) = v_f / (2 m)

v_f = 17.06 m/s

then, we get

(1.3 kg) (0 m/s) + (0.163 kg) v_0,p = [(1.3 kg) + (0.163 kg)] (17.06 m/s)

(0.163 kg) v_0,p = (24.9 kg.m/s)

v_0,p = 152.7 m/s