A bullet with a mass of 0.05 kg is fired in a straight line at a 2 kg wheel with

Question

A bullet with a mass of 0.05 kg is fired in a straight line at a 2 kg wheel with a radius of 0.3 m that is fixed to rotate about its center. Model the wheel as a solid cylinder with I=1/2 M R^2. The bullet hits the wheel at a vertical distance of 0.2 m above the center of the wheel and explodes through to the other side. After the collision, the wheel is rotating with an angular speed of 3 rad/s, while the bullet is moving at a translational speed of 50 m/s. Assume the system is isolated with negligible loss of mass in the collision. a) Find the speed of the bullet before it collided with the wheel. b) Compare the kinetic energy of the system before and after the collision. Was this an elastic or inelastic collision?

Explanation / Answer

Let

m = 0.05 kg

M = 2 kg

Moment of Inertia of wheel, (1/2)*M*R^2

= (1/2)*2*0.3^2

= 0.09 kg.m^2

after the collsion,

v = 50 m/s

w = 3 rad/s

before the collision,

u = ?

a) Apply conservation of angular momentum

m*u*r = m*v*r + I*w

0.05*u*0.2 = 0.05*50*0.2 + 0.09*3

0.01*u = 0.77

u = 0.77/0.01

= 77 m/s <<<<<<---------Answer

b) Ki = 0.5*m*u^2

= 0.5*0.05*77^2

= 148.225 J

Kf = 0.5*m*v^2 + 0.5*I*w^2

= 0.5*0.05*50^2 + 0.5*0.09*3^2

= 62.9 J

clearly, Kf < Ki

so, it is an inelastic collision.

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