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A bullet of mass m and speed v is fired at, hits and passes completely through a pendulum bob of mass M on the end of a stiff rod of negligible mass and length L. The bullet emerges with a speed of v/2 and the pendulum bob just makes it over the top of the trajectory without falling backward in its circular path. Determine an expression (V=?) for the minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle. (Use any variable or symbol stated above as necessary.)

Respuesta :

Required value of initial speed of the bullet be ( 4M/m)√(gL).

Given parameters:

Mass of the bullet =m.

Mass of the bob of the pendulum = M.

speed of the bullet before collision = v

Speed of the bullet after collision = v/2.

Length of the pendulum stiff rod = L.

Let speed transmitted to the pendulum be u.

Using principle of conservation of momentum:

mv = Mu + mv/2

⇒ Mu = mv/2

⇒ u = (m/M)v/2

We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]

To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:

u = √(4gL)

⇒  (m/M)v/2 = √(4gL)

⇒ v =( 4M/m)√(gL).

Hence, minimum required  speed of the bullet be ( 4M/m)√(gL).

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