Question

A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the stick a distance D from the middle of the stick. Before the collision, the puck was moving with speed v0 in a direction perpendicular to the stick, as indicated in the figure. The collision is completely inelastic, and the puck remains attached to the stick after the collision.
What is the angular momentum Lcm of the system before the collision, with respect to the center of mass of the final system?
Express Lcm in terms of the given variables.

Answer 1

The angular momentum of the hockey stick and puck system before collision is $m_pv_0D$.

The given parameters;

• mass of the hockey, = ms
• length of the hockey, = L
• mass of puck, = mp
• distance the puck hits the stick from the middle of the stick's length, r = D
• initial velocity of the puck, = v₀

The linear momentum of the hockey stick and puck system before collision is calculated as;

$P_i = m_s v_0_s + m_p v_0\\\\P_i = m_s(0) + m_p v_0\\\\P_i = m_pv_0$

The angular momentum of the system before collision is calculated as follow;

$L = P_i \times r\\\\L = m_pv_0 \times D\\\\L = m_pv_0D$

Thus, the angular momentum of the hockey stick and puck system before collision is $m_pv_0D$.

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Answer 2

The required angular momentum of the system before the collision is $m_{p}v_{0} D$.

Given data:

The mass of hockey stick is, $m_{s}$.

The length of hockey stick is, L.

The mass of puck is, $m_{p}$.

The hitting distance from the middle of stick is, D.

The speed of puck is, $v_{0}$.

Before collision, the total linear momentum is,

$P=m_{s}v_{0}+m_{p}v_{0}\\P=m_{s}(0)+m_{p}v_{0}\\P=m_{p}v_{0}$

And total Angular momentum before collision is,

$L=P \times D\\L=(m_{p}v_{0}) \times D\\L=m_{p}v_{0} D$

Thus, the required angular momentum of the system before the collision is $m_{p}v_{0} D$.

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