A 10 g bullet is fired horizontally into a 1 kg block of wood hanging from a 1 m long string attached to the ceiling, getting stuck in the block. Due to the impact, the block then swings on the string to a maximum angle of 60 degrees with respect to the vertical. How fast was the bullet traveling in m/s?
Details and assumptions
The acceleration of gravity is −9.8 m/ss.
You may neglect air resistance.
http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/balpen.html
i could not get there hw to put angle
That's easy. From the given conditions, height= length/2
To find the speed of the bullet, we can use the principle of conservation of energy. We know that the block is initially at rest and the bullet gets stuck in the block, causing it to swing. Therefore, the initial kinetic energy of the system (bullet + block) is zero.
The final position of the block is at a maximum angle of 60 degrees with respect to the vertical. At this point, the gravitational potential energy is at its maximum, and the kinetic energy is zero.
The total mechanical energy of the system is conserved, so we can equate the initial kinetic energy to the final potential energy:
0 = m * g * h
Where:
m = mass of the block (1 kg)
g = acceleration due to gravity (-9.8 m/ss)
h = maximum height reached by the block
To find the height, we need to consider the angle of swing. The height (h) can be calculated using the formula:
h = L * (1 - cosθ)
Where:
L = length of the string (1 m)
θ = angle of swing (60 degrees)
Converting the angle to radians:
θ (radians) = θ(degrees) * π / 180
Substituting the values into the formula:
h = 1 * (1 - cos(60 * π / 180))
Now we can calculate the height:
h ≈ 0.5 m
Plugging the values into the equation for the conservation of energy:
0 = 1 * -9.8 * 0.5
Solving for the unknown, we find:
-4.9 = -4.9 * v^2
Dividing both sides by -4.9:
1 = v^2
Taking the square root:
v ≈ 1 m/s
Therefore, the bullet was traveling at approximately 1 m/s.