A ballistic pendulum consists of a 1.25-kg block of wood that is hanging from the ceiling in such a way that when a bullet enters it, the block’s change in height can be recorded as it swings. A bullet having a mass of 6.25-grams and unknown velocity strikes the block and becomes imbedded in it. The impulse imparted to the block causes it to swing in such a way that its height increases by 7.15 cm.

1 What was the change in potential energy of the block/bullet combo after the collision?

2What was the speed of the block/bullet combo immediately after the collision (and before it began
to swing)?

3What was the speed of the bullet before entering the block of wood?

I know this is long but please anyone help me and I will pray for him for the rest of my very short life:)

change in PE=(M+m)g*height

1/2 (M+m)v^2=above PE change solve for v

m*Vbefore=(M+m)vabove
solve for Vbefore.

1. The change in potential energy of the block/bullet combo after the collision can be calculated using the equation ΔPE = mgh, where m is the mass of the block/bullet combo, g is the acceleration due to gravity, and h is the change in height. Since the mass of the block/bullet combo is 1.25 kg and the change in height is 7.15 cm (0.0715 m), we can plug in these values to find the change in potential energy.

ΔPE = (1.25 kg)(9.8 m/s^2)(0.0715 m)

Calculating this will give you the change in potential energy.

2. The speed of the block/bullet combo immediately after the collision (and before it began to swing) can be determined using the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision. Since the bullet becomes embedded in the block, the mass of the block/bullet combo after the collision is the sum of their masses. We can set up the equation as follows:

(mass of bullet)(velocity of bullet) = (mass of block/bullet combo)(velocity of combo)

Plug in the known values:

(6.25 g)(unknown velocity) = (1.25 kg + 6.25 g)(unknown velocity of combo)

Solve this equation to find the speed of the block/bullet combo immediately after the collision.

3. To calculate the speed of the bullet before entering the block of wood, we can use the principle of conservation of momentum again. Before the collision, the bullet has momentum, but the block does not. After the collision, the bullet becomes embedded in the block, and they both move as one unit. The equation can be set up as:

(mass of bullet)(velocity of bullet) = (mass of block/bullet combo)(velocity of combo)

Plug in the known values:

(6.25 g)(unknown velocity of bullet) = (1.25 kg + 6.25 g)(unknown velocity of combo)

Solve this equation to find the speed of the bullet before entering the block of wood.

Remember, clowns can be quite resourceful, but always double-check your calculations!

1. To calculate the change in potential energy, we need to determine the initial and final heights of the block/bullet combo. The change in potential energy is given by the formula:

ΔPE = m * g * Δh

where ΔPE is the change in potential energy, m is the mass of the block/bullet combo, g is the acceleration due to gravity (approximately 9.8 m/s^2), and Δh is the change in height.

Given:
m = 1.25 kg
Δh = 7.15 cm = 0.0715 m

Substituting these values into the formula:

ΔPE = 1.25 kg * 9.8 m/s^2 * 0.0715 m
= 0.8735 J

Therefore, the change in potential energy of the block/bullet combo after the collision is approximately 0.8735 Joules.

2. The speed of the block/bullet combo immediately after the collision is equal to the speed of the bullet before the collision. This is due to the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant.

The momentum of an object is given by the formula:

p = m * v

where p is the momentum, m is the mass, and v is the velocity.

Given:
m (bullet) = 6.25 grams = 0.00625 kg
m (block) = 1.25 kg

Let's assume the velocity of the bullet before the collision is v_bullet and the velocity of the block after the collision is v_block/bullet.

Then, using conservation of momentum:

(m_bullet * v_bullet) = (m_block * v_block/bullet)

0.00625 kg * v_bullet = (1.25 kg + 0.00625 kg) * v_block/bullet

Simplifying,

v_bullet = (1.25625 kg * v_block/bullet) / 0.00625 kg
v_bullet = 200.2 * v_block/bullet

Since the bullet becomes embedded in the block, the velocity of the block/bullet combo immediately after the collision is the same as the velocity of the bullet before the collision.

Therefore, the speed of the block/bullet combo immediately after the collision is approximately 200.2 times the speed of the bullet before it entered the block.

3. To find the speed of the bullet before entering the block of wood, we can use the equation for kinetic energy:

KE = 0.5 * m * v^2

where KE is the kinetic energy, m is the mass of the bullet, and v is the velocity of the bullet.

Given:
m (bullet) = 6.25 grams = 0.00625 kg

The kinetic energy of the block/bullet combo immediately after the collision is equal to the change in potential energy (ΔPE) calculated in question 1. Therefore:

KE = ΔPE

0.5 * (1.25 kg + 0.00625 kg) * v_block/bullet^2 = 0.8735 J

Simplifying,

v_block/bullet^2 = (2 * 0.8735 J) / 1.25625 kg
v_block/bullet^2 = 1.39 m^2/s^2

Taking the square root of both sides,

v_block/bullet = √(1.39 m^2/s^2)

Therefore, the speed of the bullet before entering the block of wood is approximately the square root of 1.39 m^2/s^2, which is approximately 1.18 m/s.

To solve these questions, we can use the principle of conservation of mechanical energy and momentum. Let's break down the steps to find the solutions for each question:

1. Change in Potential Energy:
The change in potential energy (PE) can be calculated using the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the change in height.
In this case, the block's mass is 1.25 kg, the change in height is 7.15 cm, which is equivalent to 0.0715 meters, and g is approximately 9.8 m/s^2.
So the change in potential energy is given by: PE = (1.25 kg) * (9.8 m/s^2) * (0.0715 m).

2. Speed of the Block/Bullet Combo after Collision:
To find the speed of the block and bullet combination after the collision, we need to use the principle of conservation of momentum.
Momentum before the collision = Momentum after the collision.
The momentum before the collision is equal to the momentum of the bullet, which is given by: momentum = mass * velocity.
The mass of the bullet is 6.25 grams, which is equal to 0.00625 kg (converting grams to kilograms).
Let's assume the final speed of the block and bullet combo after the collision is v.

Using conservation of momentum, we have:
(mass of bullet) * (velocity of bullet before) = (mass of block/bullet combo) * (velocity of block/bullet combo after)

Substituting the values, we get:
(0.00625 kg) * (velocity of bullet before) = (1.25 kg + 0.00625 kg) * v

Simplifying the equation, we can find the value of v, which represents the speed of the block/bullet combo after the collision.

3. Speed of the Bullet before Entering the Block:
To find the speed of the bullet before entering the block, we can again use the principle of conservation of momentum.
The momentum of the bullet before the collision is given by: momentum = mass * velocity.
Substituting the values, we have:
(0.00625 kg) * (velocity of bullet before) = (1.25 kg) * (velocity of block/bullet combo after)

Simplifying the equation, we can find the value of the velocity of the bullet before it entered the block.

By following these steps and substituting the given values into the equations, you should be able to find the solutions to the questions.