To test the speed of a bullet, you create a pendulum by attaching a 4.30 kg wooden block to the bottom of a 2.90 m long, 1.90 kg rod. The top of the rod is attached to a frictionless axle and is free to rotate about that point.
You fire a 10 g bullet into the block, where it sticks, and the pendulum swings out to an angle of 25.5°. What was the speed of the bullet?
(Treat the wooden block as a particle.)
speed of bullet = Sb
speed of pendulum at bottom = Sp
m = bullet mass = 0.010 Kg
initial angular momentum about ceiling hook = m v R
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mass at end after collision = 0.010 + 4.30 = 4.31 Kg
It = total moment of inertia after collision = 4.31(2.9^2)+(1/3)(1.9)(2.9)^2
w is angular velocity after collision
angular momentum after collision
It w = original ang momentum = m v R
so v = It w/ Rm
now
potential energy at top = total mass * g *height of center of mass
= (1/2) It (v^2/R^2)
height of center of mass = (1-cos25.5)(distance of center of mass from ceiling hook)
To determine the speed of the bullet, we can use the principle of conservation of mechanical energy.
In this case, the initial energy is the kinetic energy of the bullet, and the final energy is the potential energy of the pendulum at its highest point.
The final potential energy can be calculated using the equation:
PE = mgh
Where m is the mass of the block, g is the acceleration due to gravity, and h is the maximum height reached by the pendulum.
In this case, we can use the fact that the pendulum swings out to an angle of 25.5° to find h.
h = L(1 - cosθ)
Where L is the length of the pendulum and θ is the angle of swing.
Substituting the given values, we have:
h = 2.90 m (1 - cos 25.5°)
Next, we can calculate the final potential energy:
PE = (4.30 kg)(9.8 m/s^2)(2.90 m) (1 - cos 25.5°)
To determine the initial kinetic energy of the bullet, we can use the equation:
KE = (1/2)mv^2
Where m is the mass of the bullet and v is the velocity of the bullet.
We're given the mass of the bullet as 10 g, which is equivalent to 0.01 kg.
So, we have:
KE = (1/2)(0.01 kg)(v^2)
Now, since the bullet sticks to the block, the total mass after the collision is (4.30 kg + 0.01 kg) = 4.31 kg, and they move together after the collision.
According to the conservation of momentum, the momentum before the collision is equal to the momentum after the collision:
(m1)(v1) + (m2)(v2) = (m1 + m2)(v')
Where m1 and v1 are the mass and velocity of the bullet before the collision, m2 is the mass of the block, v2 is the velocity of the block after the collision, and v' is the combined velocity of the bullet and block after the collision.
Since the block is initially at rest and the bullet sticks to it,
(0.01 kg)(v1) + (4.30 kg)(0) = (0.01 kg + 4.30 kg)(v')
So, we have:
v' = (0.01 kg)(v1) / 4.31 kg
To find v' in terms of v1, we need to express the equation in terms of v' and v1:
v' = 0.01 kg * v1 / 4.31 kg
Now, we can equate the initial kinetic energy of the bullet to the final potential energy of the pendulum:
(1/2)(0.01 kg)(v^2) = (4.30 kg)(9.8 m/s^2)(2.90 m) (1 - cos 25.5°)
We have two unknowns, v and v1, but we can use the relationship between v and v1 to solve for v.
Substituting v' into the equation, we have:
(1/2)(0.01 kg)(v1^2) = (4.30 kg)(9.8 m/s^2)(2.90 m) (1 - cos 25.5°)
Simplifying this equation allows us to solve for v1, which gives us the speed of the bullet.