Angular momentum A wooden block of mass M resting on a Motio

Angular momentum. A wooden block of mass M resting on a Motionless horizontal surface is attached to a rigid rod of length l and of negligible mass, as shown in the figure to the right. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and normal to the rod with speed v hits the block and becomes embedded in it. What is the linear momentum of the bullet-block system? What is the angular momentum of the bullet-block system? Is the linear momentum conserved during the collision? Explain. Is the angular momentum conserved during the collision? Explain. What fraction of the original kinetic energy is lost in the collision?

Solution

a)

The linear momentyum of the system is conerved since there is no external force. so intial linear momentum = final linear momnetum

Inital linear momentum = mass of ball * velocyt = mv.

b) since there are no external torques acting on this system, angular momentum is conserved
the initial angular momentum with respect to the pivot point is due to the bullet\'s trajectory and is

Lbullet = mvL sin(theta) where theta is the angle between the bullet\'s velocity vector and the rod; this angle is 90 so sin 90=1

Lbullet = m v L

c) linear momentum is conserved since there is no external force.

d) since there are no external torques acting on this system, angular momentum is conserved

b)

to find KE:

the angular momentum after collision is I w
I=moment of inertia
w = angular velocity

the moment of inertia of this system is (M+m)L^2 since we can consider this as a point particle of mass m+m a distance L from the rotation axis

so we have

mvL=(m+M)L^2 w

angular velocity, w, is linked to the post collision linear velocity, V, by V=w L so w=V/L and we have

mvL = (m+M)L^2(V/L) = (m+M)VL or

V= mv/(m+M)

the ratio of KEs is

KEbefore/KE after = [1/2 mv^2]/[1/2 (m+M) {mv/(m+M)}^2] =

(m+M)/m or KE after/KE before = m/(m+M)

the amount lost is 1-m/(m+M) = M/(m+M)