A Wooden block of mass M resting on a frictionless horizontal surface is attache

Question

A Wooden block of mass M resting on a frictionless horizontal surface is attached to a rigid rod of length I 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. (a) Is the linear mementum conserved during the collision? M Explain. (b) Use censervation of momentum to calculate the final speed of the bullet/block system (c) Is the angular momentum conserved during the collision? (hint: calculate about the pivot point) (d) Calculate the final angular velocity of the bullet/block system after the collision (e) Is the linear momentum conserved after the collision? Explain. (f) Is angular momentum conserved after the collision? Explain. (g) What fraction of the original kinetic energy is lost in the collision?

Explanation / Answer

(a)The linear momentum is conserved in the collison.

The loss in momentum of the bullet is compensated by the motion of bullet block system.

b)Pi = Pf

m v = (M + m) Vf

Vf = m v (M + m)

c)Yes Angular momentum is conserved.

d)Li = I wi

Li = m wi

Lf = (M + m) l^2 wf

Li = Lf

m l^2 wi = (M + m) l^2 wf

wf = m wi/(M + m)

e)Yes the linear momentum of the system is conserved since after the collision the block bullet system has gained the velocity to compensate the loss in velocity of bullet.

f)Yes the angular momentum of the system is conserved.

g)K(lost) = Kf - Ki

K(lost) = 1/2 I wf^2 - 1/2 m v^2

I = (M + m) l^2 ; I w^2 = (M + m) vf^2

Li = Lf

m v l = (M + m) vf l

k(loss) = M/(M + m) Ki

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