Question

A long thin bar (length L 18 cm, mass 1.8 kg) of uniform density is placed upon a horizontal, frictionless surface. A small rubber puck (mass 250 g) slides towards the bar with a speed (2 m/s) directed perpendicular to the bar. It collides perfectly elastically with the bar at a distance (d) from the center of mass of the bar in such a way that the puck rebounds with a velocity (1 m/s). a) What is the value of 'd'? b) Calculate the bar's center of mass velocity after the collision. c) Calculate the bar's angular velocity after the collision. BONUS: Repeat the problem, but this time the bar is a square m bar with width (w 4 cm). The puck has a radius (r 4 cm) C and the collision is perfectly inelastic.

A long thin bar (length L = 18 cm, mass 1.8 kg) of uniform density is placed upon a horizontal, frictionless surface. A small rubber puck (mass 250 g) slides towards the bar with a speed (2 m/s) directed perpendicular to the bar. It collides perfectly elastically with the bar at a distance (d) from the center of mass of the bar in such a way that the puck rebounds with a velocity (1 m/s).

a) What is the value of ‘d’?

b) Calculate the bar’s center of mass velocity after the collision.

c) Calculate the bar’s angular velocity after the collision.

BONUS: Repeat the problem, but this time the bar is a square bar with width (w = 4 cm). The puck has a radius (r = 4 cm) and the collision is perfectly inelastic.

Answer 1

The moment of inertia of the rod I = ML^2/12 = 1.8*0.18^2/12 = 0.00486 Kg m^2

By conservation of linear momentum,

m1u1 + 0 = m1v1 + m2v2

0.25*2 = -0.25*1 + 1.8v2

=> v2 = 0.417 m/s

Therefore angular velocity of bar is w = v/r = 0.417/0.09 (since r = L/2)

w = 4.629 rad/s

Now by conservation of angular momentum

m1u1d + 0 = m1v1d + Iw

0.25*2*d = -0.25*1*d + 0.00486*4.629

d = 0.0299 m = 2.99 cm = 3 cm

b) The bar's center of mass velocity is as calculated above i.e v = 0.417 m/s

c) The bar's angular velocity is as calculated above i.e w = 4.629 rad/s