A 15g bullet is fired horizontally into a block of wood with a mass of 2.5 k and embedded in the block. Initially the block of wood hangs vertically and the impact causes the block o swing so that its center of mass rises 15cm. find the velocity of the bullet just before the impact.
increase of potential energy of block = m g h
= (2.515)(9.81)(.15) = 3.70 Joules
so
(1/2) m v^2 = 3.7 Joules at bottom
(1/2)(2.515)v^2 = 3.7
v = 1.72 m/s at bottom after collision
momentum after collision
= 2.515 * 1.72 = 4.31 g m/s
momentum before collision = .015 v
so
v = 4.31 / .015 = 288 m/s
He knows the mass of the bullet is 85 g. He fires the gun into a 1.3 kg block and finds the speed of the block/bullet combo to be 67 m/s. How fast was the bullet traveling before it hit the block?
To find the velocity of the bullet just before the impact, we can use the principle of conservation of momentum.
1. First, we need to find the initial velocity of the block of wood before the impact.
- The block of wood starts from rest, so its initial velocity (u) is 0 m/s.
2. Next, we'll determine the final velocity of the block of wood and bullet together after the impact, which is also the velocity of the bullet just before the impact.
- The mass of the bullet, m1 = 15 g = 0.015 kg
- The mass of the block of wood, m2 = 2.5 kg
- Let v be the final velocity of the block of wood and bullet together.
3. Using the principle of conservation of momentum, we have:
- (m1 * u1) + (m2 * u2) = (m1 * v1) + (m2 * v2)
- As the block starts from rest, u2 (initial velocity of the block) is 0 m/s.
- Also, the bullet is embedded in the block after the impact, so the final velocity of the bullet (v1) is 0 m/s.
- The equation simplifies to: (m1 * u1) = (m2 * v2)
4. Now, we'll calculate the value of v2 (final velocity of the block of wood and bullet together).
- The change in height, h = 15 cm = 0.15 m (converted to meters)
- The gravitational acceleration, g = 9.8 m/s^2
- Using the conservation of mechanical energy principle, the potential energy gained by the system (block of wood and bullet) is equal to the loss in kinetic energy due to the rise in height.
- m2 * g * h = (1/2) * (m1 + m2) * v2^2
- Plugging in the values, we get: 2.5 * 9.8 * 0.15 = (1/2) * (0.015 + 2.5) * v2^2
- Solving for v2, we find v2 ≈ 2.936 m/s
So, the velocity of the bullet just before the impact is approximately 2.936 m/s.
To find the velocity of the bullet just before the impact, we can use the principle of conservation of momentum. This principle states that the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.
Let's break down the problem into parts and determine the initial momentum and final momentum.
1. Initial momentum: The bullet is fired horizontally, so its initial momentum is given by the equation:
momentum of the bullet = mass of the bullet × velocity of the bullet
Given:
mass of the bullet (m1) = 15 g = 0.015 kg (convert grams to kilograms)
velocity of the bullet (v1) = unknown
Therefore, the initial momentum (p1) of the bullet is: p1 = m1 × v1
2. Final momentum: After the collision, the bullet and the block of wood move together. Since the bullet gets embedded in the block, their combined mass becomes the mass of the block plus the bullet. The final momentum (p2) can be expressed as:
p2 = (mass of the block + mass of the bullet) × velocity of the block and bullet (v2)
Given:
mass of the block (m2) = 2.5 kg
velocity of the block and bullet (v2) = unknown
Now, the principle of conservation of momentum states that p1 equals p2:
m1 × v1 = (m2 + m1) × v2
Plugging in the known values:
0.015 kg × v1 = (2.5 kg + 0.015 kg) × v2
v1 = (2.515 kg) × v2
The block of wood swings up, which means it gains potential energy. Using the formula for potential energy, we can relate the change in potential energy to the work done by the bullet:
change in potential energy = work done by the bullet
The change in potential energy is given by:
change in potential energy = mass × gravity × height
Given:
mass of the block (m2) = 2.5 kg
gravity (g) = 9.8 m/s^2
height (h) = 15 cm = 0.15 m (convert cm to meters)
Substituting the values:
change in potential energy = 2.5 kg × 9.8 m/s^2 × 0.15 m
Since the work done by the bullet is equal to the change in potential energy, we can express it as:
work done by the bullet = force × displacement
The force can be determined using Newton's Second Law:
force = mass × acceleration
Given:
mass of the bullet (m1) = 0.015 kg
acceleration (a) = unknown
We can assume that the acceleration (a) is constant throughout the motion.
The force can be written as:
force = m1 × a
Since the displacement (displacement) of the block is vertical and perpendicular to the force applied, we can write:
work done by the bullet = force × displacement = m1 × a × displacement
Comparing the work done by the bullet equation to the change in potential energy equation, we can equate them since they represent the same physical effect:
m1 × a × displacement = 2.5 kg × 9.8 m/s^2 × 0.15 m
Now we have two equations:
1. 0.015 kg × v1 = (2.515 kg) × v2
2. 0.015 kg × a × displacement = 2.5 kg × 9.8 m/s^2 × 0.15 m
From these two equations, we can solve for v1 and find the velocity of the bullet just before the impact.