An 8.00 g bullet is fired into a 180 g block that is initially at rest at the edge of a table of 1.00 m height. The bullet remains in the block, and after the impact the block lands d = 2.2 m from the bottom of the table. Determine the initial speed of the bullet.
initial momentum = final momentum
.008 * V + .18 * 0 = .188 U
where V is initial bullet speed and U is horizontal speed off the table.
Now get U
t is fall time
horizontal distance = U t = 2.2
vertical distance = 4.9 t^2 = 1
t= sqrt (1/4.9)
solve that for t
U = 2.2 / t
V = (.188/.008) U
420
To determine the initial speed of the bullet, we can use the principle of conservation of momentum. In this case, we assume that no external forces act on the system (the bullet and the block) after the collision.
The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.
Let's break down the problem step by step:
Step 1: Calculate the momentum of the bullet before the collision.
- Mass of the bullet (m1) = 8.00 g = 0.008 kg (converted to kilograms)
- Velocity of the bullet before the collision (v1) = ?
- Momentum of the bullet before the collision (p1) = m1 * v1
Step 2: Calculate the momentum of the block before the collision.
- Mass of the block (m2) = 180 g = 0.180 kg (converted to kilograms)
- Since the block is initially at rest, the velocity of the block before the collision (v2) = 0
- Momentum of the block before the collision (p2) = m2 * v2
Step 3: Calculate the momentum of the bullet and the block after the collision.
- Mass of the bullet and block combined (m3) = m1 + m2
- Combined velocity after the collision (v3) = ?
- Momentum of the bullet and block after the collision (p3) = m3 * v3
Since we know that the total momentum before the collision is equal to the total momentum after the collision, we have: p1 + p2 = p3.
Using the equations from Step 1, Step 2, and Step 3, we can write the equation as:
(m1 * v1) + (m2 * v2) = (m3 * v3)
Plugging in the values we know:
(0.008 kg * v1) + (0.180 kg * 0) = ((0.008 kg + 0.180 kg) * v3)
Simplifying:
0.008 kg * v1 = 0.188 kg * v3
Now, we can solve for the initial velocity of the bullet (v1) by rearranging the equation:
v1 = (0.188 kg * v3) / 0.008 kg
Next, we need to determine the final velocity (v3). We can do this using the equation of motion in the vertical direction.
Step 4: Calculate the final velocity of the bullet and the block after falling a distance of 2.2 m.
- Height of the table (h) = 1.00 m
- Distance from the bottom of the table (d) = 2.2 m
- Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no air resistance)
- Initial velocity of the bullet and block after the collision (u) = 0 (since they are dropped)
- Final velocity of the bullet and block after falling (v) = ?
Using the equation of motion: v^2 = u^2 + 2g(d - h)
Plugging in the values:
v^2 = 0 + 2 * 9.8 m/s^2 * (2.2 m - 1.00 m)
Simplifying:
v^2 = 2 * 9.8 m/s^2 * 1.2 m
Now, we can solve for the final velocity (v) by taking the square root of both sides:
v = √(2 * 9.8 m/s^2 * 1.2 m)
To get the initial velocity (v1), we substitute the value of the final velocity (v) into the equation for v1 calculated in Step 3:
v1 = (0.188 kg * v) / 0.008 kg
Substituting the values and calculating the final result will give us the initial speed of the bullet.