A stick of mass 1.3 kg and length 0.4 m is lying at rest on ice. A small object

Question

A stick of mass 1.3 kg and length 0.4 m is lying at rest on ice. A small object with mass 0.06 kg traveling at high speed of 200 m/s strikes the stick a distance 0.2 m from the center and bounces off with spedd 60 m/s, the motion is in the xy plane and the z axis points upward from the ice. The magnitudes of the initial and final angles to the x axis of the small mass's velocity are ?i = 26 degree and ?f =82 degrees. with the moment of inertia about the center of a uniform stick and mass M and length L is I = ML

Explanation / Answer

Momentum along the x axis

m vi cos(theta_i) = m vf cos(theta_f) + M V cos(angle)

Momentum along the y axis

m vi sin(theta_i) = m vf sin(theta_f) + M V sin(angle)

where m, vi, vf, theta_i, theta_f, and M are known. So you have two equations and two unknowns (V and alpha) where V is the velocity of the stick center of mass after collision and alpha is the angle of this velocity with respect to the x-axis.

Now, conservation of angular momentum

m d vi cos(theta_i) = m d vf cos(theta_f) + I w

where I is the moment of inertia of the stick, which is given by

I = 1/12 M L^2

from the equation above, everything is known except w (angular velocity of the stick after the collision) so you can solve for that.

From conservation of energy we have

1/2 m vi^2 = 1/2 m vf^2 + 1/2 M V^2 + 1/2 I w^2 + thermal energy dissipated

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