To solve this problem we will apply the concepts related to the conservation of momentum. Momentum can be defined as the product between mass and velocity. We will depart to facilitate the understanding of the demonstration, considering the initial and final momentum separately, but for conservation, they will be later matched. Thus we will obtain the value of the mass. Our values will be defined as
Initial momentum will be
After collision
Final momentum
From conservation of momentum
Replacing,
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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Answer:
Explanation:
ASSUMING that block = sled AND that the rope is parallel to the slope.
The force acting parallel due to the weight is
13.6(9.81)sin35.5 = 77.475 N
The maximum friction force is
(0.45)13.6(9.81)cos35.5 = 48.877 N
If rope tension is T
77.475 - 48.877 < T < 77.475 + 48.877
28.6 N < T < 126 N
28.6 N will occur if the block is on the verge of sliding downhill
126 N will occur if the block is on the verge of sliding uphill
Could be any value between them.
Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision though. And that is because it is converted to another form of energy
