Question

A rigid uniform rod of length 90.0 cm and mass 35.0 kg laying on a horizontal frictionless surface is free to rotate on that surface about an axle through its center. A small puck slides into it at an angle of 90 degrees to its surface at a distance of 30.0 cm from the pivot point at a speed of 20.0 m/s. It rebounds with a speed of 16.0 m/s along the same line. If the angular speed of the rod immediately after the collision is 1.14 rad/s, what was the mass of the puck

Answer 1

m = 0.217 kg

Step-by-step explanation:

We can solve this exercise using the conservation of angular momentum. For this the system is formed by the bar and the disk, so that the forces during the crash have been internal and the angular momentum is preserved

initial angular mount. Before impact

L₀ = L_bar + L_ disk

L₀ = I_bar w₀ + m r v₀

final angular momentum. Right after the crash

`L_(f)` = I_bar wf = m r v_{f}

The moment of inertia of a bar that rotates at its ends is

I_bar = 1/12 M L

how the angular momentum is conserved

L₀ = L_{f}

I_barr w₀ + mr v₀ = I_barr w_{f} + m r v_{f}

I_bar (w₀- w_{f}) = m r (v₀- v_{f})) r

m = I_bar (w₀ - w_{f}) / r (v₀ -v_{f})

m = 1/12 M L (w₀ -w_{f} ) / r (v₀ -v_{f})

in the exercise it indicates that the initial speed of the disc is v₀ = 20 m / s and its final speed is v_{f} = -16 m / s, the negative sign is because the disc recoils

we calculate

m = 1/12 35 0.90 (0 + 1.14) / [0.30 (30- (-16))]

m = 0.217 kg