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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The linear velocity of a rotating object is the product of the angular velocity and the radius of the circular motion. Angular velocity is the rate of the change of angular displacement of a body that is in a circular motion. It is a vector quantity so it consists of a magnitude and direction. From the problem, the angular velocity is 5.9 rad per second and the radius is given as 12 centimeters. We calculate as follows:
Linear velocity = angular velocity (radius)
Linear velocity = 5.9 (12 ) = 70.8 cm / s
The linear velocity of the body in motion is 70.8 centimeters per second or 0.708 meters per second.
