A 3.30-g bullet, traveling at a speed of 492 m/s, strikes the wooden block of a ballistic pendulum, such as that in Figure 7.14. The block has a mass of 232 g. (a) Find the speed of the bullet/block combination immediately after the collision. (b) How high does the combination rise above its initial position?
1) Use the law of conservation of momentum to get the bullet_block velocity after collision, Vfinal.
3.30*492 = Vfinal*(3.30 + 232)
2) Use the law of conservation of energy to calculate how far up the bullet and block rise, H
g H = Vfinal^2/2
3) Show your work if further assistance is required
To solve this problem, we can use the principles of conservation of momentum and conservation of energy.
(a) Find the speed of the bullet/block combination immediately after the collision:
Step 1: First, calculate the momentum of the bullet before the collision.
Momentum (p) is given by the product of mass (m) and velocity (v):
momentum = mass × velocity
Given:
Mass of the bullet (m1) = 3.30 g = 0.0033 kg
Velocity of the bullet (v1) = 492 m/s
So, the momentum of the bullet before the collision is:
p1 = m1 * v1 = 0.0033 kg * 492 m/s
Step 2: Next, calculate the momentum of the wooden block before the collision.
Mass of the wooden block (m2) = 232 g = 0.232 kg
Velocity of the wooden block (v2) before the collision is assumed to be zero, as it is initially at rest.
So, the momentum of the wooden block before the collision is:
p2 = m2 * v2 = 0.232 kg * 0 m/s = 0 kg·m/s
Step 3: Apply the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
p1 + p2 = p'1 + p'2
Since the wooden block is initially at rest, its momentum before and after the collision is zero:
p2 = p'2 = 0 kg·m/s
Therefore, the equation becomes:
p1 + 0 = p'1 + 0
p1 = p'1
So, the momentum of the bullet after the collision is the same as its momentum before the collision.
Step 4: Calculate the speed of the bullet/block combination immediately after the collision.
The momentum of the bullet/block combination is given by:
p' = m' * v'
Where m' is the combined mass of the bullet and block, and v' is the combined velocity.
The mass of the bullet/block combination (m') is the sum of the masses of the bullet (m1) and the wooden block (m2):
m' = m1 + m2
Plugging in the values, we have:
m' = 0.0033 kg + 0.232 kg
Now, we can substitute this into the momentum equation to solve for the velocity (v') of the bullet/block combination after the collision:
p1 = m1 * v1 = m' * v'
Substituting the values, we get:
0.0033 kg * 492 m/s = (0.0033 kg + 0.232 kg) * v'
Simplifying the equation:
0.0033 kg * 492 m/s = 0.2353 kg * v'
Now, solve for v':
v' = (0.0033 kg * 492 m/s) / (0.2353 kg)
Calculating the value of v', we get:
v' ≈ 6.9 m/s
Therefore, the speed of the bullet/block combination immediately after the collision is approximately 6.9 m/s.
(b) How high does the combination rise above its initial position:
To find the height the combination rises, we can use the principle of conservation of energy.
Step 1: Calculate the initial kinetic energy of the bullet:
The initial kinetic energy (KE) of the bullet is given by the formula:
KE = (1/2) * m1 * v1^2
Substituting the given values:
KE = (1/2) * 0.0033 kg * (492 m/s)^2
Calculating the value of KE, we get:
KE ≈ 402.5664 J
Step 2: Calculate the potential energy gained by the combination:
The potential energy gained (PE) by the combination is given by the formula:
PE = m' * g * h
Where m' is the combined mass of the bullet and block, g is the acceleration due to gravity, and h is the height gained.
Given:
m' ≈ 0.0033 kg + 0.232 kg
g = 9.8 m/s^2
Substituting the values, we have:
PE = (0.0033 kg + 0.232 kg) * 9.8 m/s^2 * h
Step 3: Apply the principle of conservation of energy.
According to the principle of conservation of energy, the initial kinetic energy is equal to the potential energy gained:
KE = PE
Substituting the values, we get:
402.5664 J = (0.0033 kg + 0.232 kg) * 9.8 m/s^2 * h
Now, solve for h:
h = 402.5664 J / [(0.0033 kg + 0.232 kg) * 9.8 m/s^2]
Calculating the value of h, we get:
h ≈ 1.71 m
Therefore, the combination rises approximately 1.71 meters above its initial position.
To find the speed of the bullet/block combination immediately after the collision, we can use the principle of conservation of linear momentum. According to this principle, the total linear momentum before the collision is equal to the total linear momentum after the collision.
The linear momentum of an object is given by the product of its mass and velocity: momentum = mass × velocity.
Before the collision, the bullet's linear momentum is given by the product of its mass (3.30 g) and its velocity (492 m/s). However, we need to convert the mass to kilograms by dividing it by 1000: 3.30 g = 0.0033 kg. So, the initial linear momentum of the bullet is 0.0033 kg × 492 m/s.
After the collision, the bullet and the block move together as a single system. Let's assume their final velocity (v) and calculate their combined linear momentum. The mass of the block is 232 g, which is equal to 0.232 kg. So, the combined linear momentum of the bullet/block combination is (0.0033 kg + 0.232 kg) × v.
According to the principle of conservation of linear momentum, these two linear momenta should be equal:
0.0033 kg × 492 m/s = (0.0033 kg + 0.232 kg) × v
Now, we can solve this equation for v to find the speed of the bullet/block combination immediately after the collision.
Moving on to part (b) of the question, to determine how high the combination rises above its initial position, we can use the principle of conservation of mechanical energy. The mechanical energy of a system is conserved if there are no external forces doing work on it.
In this case, the only significant external force is gravity, which is a conservative force. Therefore, the initial mechanical energy of the system (bullet/block combination) is equal to its final mechanical energy.
The initial mechanical energy is given by the sum of the kinetic energy of the bullet and the potential energy of the block (both with respect to the initial reference point). The kinetic energy is given by the equation K.E. = 0.5 × mass × velocity^2, where the mass is in kilograms and the velocity is in meters per second.
The potential energy of the block is given by the equation P.E. = mass × g × height, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and the height is the vertical distance above the initial reference point.
After the collision, the final mechanical energy is equal to the potential energy of the block at its maximum height, as the kinetic energy of the bullet/block combination is zero at its highest point.
So, equating the initial mechanical energy to the final mechanical energy, we can solve for the height.
Note: Figure 7.14 was referenced in the question, but since it was not provided, the specific details of the ballistic pendulum's design cannot be considered in this explanation.