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).
Learn more about speed here:
brainly.com/question/28224010
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<em>Answer:</em>
<em>r=x+y</em>
<em>sorry if its not correct you can delete if you want.</em>
I found the answer sheet online for you
10 Km.
S= Speed
D= distance
T= time
S= d/t
but since you are solving for "d" the equation is d=st so you plug in 10 km/h for speed and 2.1 hours for time and just multiply them. The hours cancel out so you are left with 10km.
Then the tangent of angle-Θ is (Ay / Ax).
