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2 months ago

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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.

1 answer:

Sav [3.1K]2 months ago

5 0

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Provided information

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick prior to the impact)

xis = D (initial position of the center of mass of the hockey stick before impact)

mp = mass of the puck

uip = v₀ (initial velocity of the puck before impact)

xip = 0 (initial position of the center of mass of the puck before impact)

By applying

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒ Ycm = (ms*D + mp*0) / (ms + mp)

⇒ Ycm = (ms*D) / (ms + mp)

Now, using the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

we arrive at

L = mp*v₀*(ms*D) / (ms + mp)

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