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A projectile of mass m moves to the right with a speed v i. The projectile strik

ID: 1704332 • Letter: A

Question

A projectile of mass m moves to the right with a speed vi. The projectile strikes and sticks to the end of a stationary rod of mass M, lenfth d, pivoted though O. we wish to find the fractional change of kinetic energy in the system due to the collisions.

a) What is the appropriate analysis model to describe the prijectile and the rod?

b) What is teh angular momentum of the system before the collison aboutan axis through O?

c) What is the moment of inertia of the system about an axis through O after the projectile sticks to the rod?

d) If the angular speed of the system after the collision is w, what is the angular momentium of the system after the collision?

e) Find the angulat speed w after the collision in terms of the given quanities.

f) What is the kinetic energy of the system before the collision?

g) What is the kinetic energy of the system after the collision?

h) Determine the fractional change of kinetic energy due to the collision?

Explanation / Answer

A)

As the collision is inelastic, conservation of angular momentum will be the appropriate analysis model to describe the projectile and the rod

B)

As the rod is at rest, the angular momentum of the rod before collision is zero

The angular momentum of mass m moving with speed vi at a distance d/2 from the pivot will be L = mvid/2

The angular momentum of the system before collision is

Li= mvid/2

C)

The moment of inertia of the rod through the axis passing through O is Irod = Md2/12

The moment of inertia of the mass m is Imass = md2/4

The moment of inertia of the system is I = Md2/12 + md2/4 = (M+3m)d2/12

D)

is the angular speed of the system after collision

Then the final angular speed of the system will be Lf = I

Lf = [(M+3m)d2/12]

E)

The angular speed of the system after the collision is

= 12Lf /(M+3m)d2

From conservation of angular momentum, Li = Lf

= 12Li /[(M+3m)d2

= 12[mvid/2]/[(M+3m)d2

= [6m/(M+3m)]vi/d

F)

The kinetic energy of the system before collision is

KEi = 0.5mvi2   (The rod is at rest)

G)

The kinetic energy of the system after collision is

KEf = 0.5I2

KEf = 0.5*[(M+3m)d2/12]*[ [6m/(M+3m)]vi/d]2

KEf = 3/2 * m2vi2/(M+3m)

KEf = 3mKEi /(M+3m)

Where KEi = 0.5mvi2

H)

The fractional change of kinetic energy due to the collision is

KEf /KEi = 3m/(M+3m)