A bullet of mass 5.15 g is fired horizontally into a 2.74 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.185. The bullet comes to rest in the block, which moves 1.84 m.

(a) What is the speed of the block immediately after the bullet comes to rest within it?
(b) At what speed is the bullet fired?

total mass = 5.15 + 2.74 = 2.74515 kg

momentum before
.00515 V
momentum after = .00515 V = 2.74515 Vi

work done on system by friction = mu M d
= .185 * 2.74515 * 1.84
= .934 Joules
that was the Ke of the system at Vi
(1/2)(2.74515)Vi^2 = .934 Joules
Vi = .825 m/s part a
V = .825 ( 2.74515/.00515) = 440 m/s

work done on system by friction = mu M g d

= .185 * 2.74515 * 1.84 * 9.8
= 9.15 Joules
that was the Ke of the system at Vi
(1/2)(2.74515)Vi^2 = 9.15 Joules
Vi =2.58 m/s part a
V = 2.58 ( 2.74515/.00515) = 1376 m/s

To solve this problem, we can apply the principles of conservation of momentum and conservation of energy.

(a) To find the speed of the block immediately after the bullet comes to rest within it, we need to use the principle of conservation of momentum. This principle states that the total momentum before an event is equal to the total momentum after the event.

Let's denote the initial speed of the bullet as v1 and the final speed of the combined block and bullet as v2. The momentum before the event is given by the product of the mass and the initial velocity of the bullet, while the momentum after the event is given by the product of the total mass (block + bullet) and the final velocity.

Momentum before = Momentum after
(mass of bullet) x (initial speed of bullet) = (mass of block + bullet) x (final speed of block + bullet)

Converting the given masses and speed to SI units:
mass of bullet = 5.15 g = 5.15 x 10^-3 kg
mass of block = 2.74 kg

Let's assume the final speed of the bullet inside the block is zero, and the final speed of the block is v2.

(5.15 x 10^-3 kg) x v1 = (2.74 kg + 5.15 x 10^-3 kg) x v2
v1 = (2.74 kg + 5.15 x 10^-3 kg) x v2 / (5.15 x 10^-3 kg)

Let's substitute the given values into the equation and solve for v1:

v1 = (2.74 kg + 5.15 x 10^-3 kg) x 0 / (5.15 x 10^-3 kg)
v1 = 0 m/s

Therefore, the speed of the block immediately after the bullet comes to rest within it is 0 m/s.

(b) To find the speed at which the bullet is fired, we can use the principle of conservation of energy. The initial kinetic energy of the bullet is equal to the work done against friction and the final kinetic energy of the block-bullet system.

The initial kinetic energy of the bullet can be given by the equation:
0.5 x (mass of bullet) x (initial speed of bullet)^2

The work done against friction can be given by the equation:
(work done) = (force of friction) x (displacement)
(work done) = (coefficient of kinetic friction) x (normal force) x (displacement)
(normal force) = (mass of block) x (acceleration due to gravity)
= (2.74 kg) x (9.8 m/s^2)

The final kinetic energy of the block-bullet system can be given by the equation:
0.5 x (mass of block + bullet) x (final speed of block + bullet)^2

Setting the initial kinetic energy equal to the work done against friction plus the final kinetic energy:

0.5 x (mass of bullet) x (initial speed of bullet)^2 = (coefficient of kinetic friction) x (mass of block) x (acceleration due to gravity) x (displacement) + 0.5 x (mass of block + bullet) x (final speed of block + bullet)^2

We are given the values for the mass of the bullet, mass of the block, coefficient of kinetic friction, and the displacement. We need to solve for the initial speed of the bullet (v1).

Let's substitute the given values and solve for v1:

0.5 x (5.15 x 10^-3 kg) x v1^2 = (0.185) x (2.74 kg) x (9.8 m/s^2) x (1.84 m) + 0.5 x (2.74 kg + 5.15 x 10^-3 kg) x v2^2

Since we found in part (a) that v2 is equal to 0 m/s:

0.5 x (5.15 x 10^-3 kg) x v1^2 = (0.185) x (2.74 kg) x (9.8 m/s^2) x (1.84 m)

Now, let's solve for v1:

v1^2 = [(0.185) x (2.74 kg) x (9.8 m/s^2) x (1.84 m)] / (0.5 x (5.15 x 10^-3 kg))
v1 = sqrt([(0.185) x (2.74 kg) x (9.8 m/s^2) x (1.84 m)] / (0.5 x (5.15 x 10^-3 kg)))

Using a calculator to evaluate the right side of the equation:

v1 ≈ 220.7 m/s

Therefore, the speed at which the bullet is fired is approximately 220.7 m/s.

To solve this problem, we need to apply the principles of conservation of momentum and conservation of kinetic energy.

(a) What is the speed of the block immediately after the bullet comes to rest within it?

1. Calculate the momentum of the bullet before collision:
momentum = mass * velocity
mass of the bullet = 5.15 g = 0.00515 kg (converted to kg)
velocity of the bullet = unknown (let's call it v)
momentum = 0.00515 kg * v

2. Calculate the momentum of the block after collision:
momentum = mass * velocity
mass of the block = 2.74 kg
velocity of the block after the bullet comes to rest = unknown (let's call it V)
momentum = 2.74 kg * V

According to the law of conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Therefore, we can set up the equation:

0.00515 kg * v = 2.74 kg * V

3. Solve for V:
V = (0.00515 kg * v) / 2.74 kg
= (v/532.4) m/s

Since we know that the block moves 1.84 m, we can determine the time it took for the bullet to come to rest inside the block. We can use the equation of motion:

Distance = (initial velocity * time) + (0.5 * acceleration * time^2)

In this case, there are two forces acting on the block:
- The bullet's momentum gets transferred to the block, causing it to move
- The kinetic friction force slows down the block

4. Calculate the acceleration of the block:
The net force acting on the block is equal to the frictional force. We can calculate it using the formula:

frictional force = coefficient of kinetic friction * normal force

The normal force is equal to the weight of the block:
normal force = mass of the block * acceleration due to gravity

Since the block is on a horizontal surface, the normal force cancels out the gravitational force acting on the block, leaving only the frictional force:

frictional force = coefficient of kinetic friction * (mass of the block * acceleration due to gravity)

Now, we can calculate the acceleration of the block using Newton's second law:

net force = mass * acceleration
frictional force = mass of the block * acceleration

Substituting the values we have:

coefficient of kinetic friction * (mass of the block * acceleration due to gravity) = 2.74 kg * acceleration

Solving for acceleration:

acceleration = (coefficient of kinetic friction * acceleration due to gravity)

5. Calculate the time it took for the bullet to come to rest in the block:
Using the equation of motion, rearranged for time:

Distance = (initial velocity * time) + (0.5 * acceleration * time^2)

1.84 m = (v/532.4) m/s * time + (0.5 * (coefficient of kinetic friction * acceleration due to gravity) m/s^2 * time^2)

This is a quadratic equation in terms of time. Use the quadratic formula to solve for time:

time = (-b ± √(b^2 - 4ac)) / 2a

Where:
a = 0.5 * (coefficient of kinetic friction * acceleration due to gravity) m/s^2
b = (v/532.4) m/s
c = -1.84 m

6. Calculate the speed of the block after the bullet comes to rest within it:
Since we have calculated the time, we can use it with the acceleration equation to find the final velocity of the block:

velocity = (initial velocity) + (acceleration * time)

In this case, the initial velocity is v and the acceleration is given by the equation:

acceleration = coefficient of kinetic friction * acceleration due to gravity

Now, substitute the values we have:

velocity = (v/532.4) m/s - (coefficient of kinetic friction * acceleration due to gravity) m/s^2 * time

7. Substitute the calculated time into the above equation to find the final velocity (V) of the block:
V = (v/532.4) m/s - (coefficient of kinetic friction * acceleration due to gravity) m/s^2 * calculated time

(b) At what speed is the bullet fired?

We know that the bullet comes to rest within the block, so the final velocity of the bullet is 0 m/s. Therefore, we can set up the equation:

(v/532.4) m/s - (coefficient of kinetic friction * acceleration due to gravity) m/s^2 * calculated time = 0 m/s

Solve the equation for v:

(v/532.4) m/s = (coefficient of kinetic friction * acceleration due to gravity) m/s^2 * calculated time

v = (coefficient of kinetic friction * acceleration due to gravity) m/s^2 * calculated time * 532.4 m/s

Now, substitute the calculated values of the coefficient of kinetic friction, acceleration due to gravity, and calculated time into the above equation to find the speed at which the bullet is fired (v).