A block of mass M = 1 kg is attached to a massless rope of length 1 m which is 1 = initially at rest. A bullet with a mass of m 10 gm and a speed of Vo = 400 m/s strikes the block horizontally. The bullet embeds itself there and the whole system swings. Find the maximum angle of the swinging.
Answers & Comments
Answer:85.5
Explanation:
To find the maximum angle of the swinging, we need to use the conservation of mechanical energy.
The initial mechanical energy of the system is the kinetic energy of the bullet:
Ei = 1/2 * m * Vo^2
Where m is the mass of the bullet and Vo is the speed of the bullet.
The final mechanical energy of the system is the potential energy of the swinging block:
Ef = mgh
Where m is the mass of the block, g is the acceleration due to gravity and h is the maximum height of the swinging block.
Since the total mechanical energy is conserved, we can set these two equations equal to each other:
1/2 * m * Vo^2 = mgh
Solving for h, we get:
h = Vo^2 / (2g)
Now, to find the maximum angle of the swinging, we can use the length of the rope and the height of the swinging block.
The maximum angle of the swinging is given by:
theta = atan(h/L)
Where L is the length of the rope.
Substituting the values, we get:
theta = atan(Vo^2 / (2gL))
Where g = 9.8 m/s^2 and L = 1 m
theta = atan(400^2 / (29.81))
theta = atan(160000/19.6)
theta = atan(8204.16)
theta = 85.5 degrees
Therefore, the maximum angle of the swinging is 85.5 degrees.