A 2.86-g bullet, traveling at a speed of 452 m/s, strikes the wooden block of a ballistic pendulum, such as that in Figure 7.14. The block has a mass of 230 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?
To find the answers to these questions, we can make use of the principle of conservation of momentum and conservation of mechanical energy.
(a) Find the speed of the bullet/block combination immediately after the collision:
1. Start by calculating the momentum of the bullet before the collision. Momentum is given by the equation p = mv, where p is momentum, m is mass, and v is velocity. In this case, the mass of the bullet is 2.86 g, which is equivalent to 0.00286 kg. The velocity of the bullet is 452 m/s. Therefore, the initial momentum of the bullet is:
p_bullet = (0.00286 kg) * (452 m/s) = 1.29472 kg·m/s
2. Next, calculate the momentum of the block before the collision. The mass of the block is 230 g, or 0.23 kg. Since the block is initially at rest, the momentum of the block is:
p_block = 0 kg·m/s
3. Conservation of momentum tells us that the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can write:
p_bullet + p_block = p_combined
1.29472 kg·m/s + 0 kg·m/s = p_combined
p_combined = 1.29472 kg·m/s
4. Assuming that there is no external force acting on the system after the collision, the total momentum after the collision is also equal to the momentum of the combined bullet and block. Therefore:
p_combined = (m_bullet + m_block) * v_combined
1.29472 kg·m/s = (0.00286 kg + 0.23 kg) * v_combined
5. Rearrange the equation and solve for v_combined:
v_combined = 1.29472 kg·m/s / (0.00286 kg + 0.23 kg) ≈ 5 m/s
Therefore, the speed of the bullet/block combination immediately after the collision is approximately 5 m/s.
(b) How high does the combination rise above its initial position:
1. To find the height reached by the combination, we can use the principle of conservation of mechanical energy. At the maximum height, the combination has no kinetic energy and only potential energy due to its height above the initial position.
2. The initial kinetic energy of the bullet/block combination is:
KE_initial = 1/2 * (m_bullet + m_block) * v_combined^2
KE_initial = 1/2 * (0.00286 kg + 0.23 kg) * (5 m/s)^2
3. The final potential energy of the combination at its maximum height is given by:
PE_final = (m_bullet + m_block) * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the maximum height.
4. Since mechanical energy is conserved, we can equate the initial kinetic energy to the final potential energy:
KE_initial = PE_final
1/2 * (0.00286 kg + 0.23 kg) * (5 m/s)^2 = (0.00286 kg + 0.23 kg) * 9.8 m/s^2 * h
5. Cancel out the masses and solve for h:
(1/2) * (5 m/s)^2 = 9.8 m/s^2 * h
h = [(1/2) * (5 m/s)^2] / (9.8 m/s^2)
h ≈ 1.275 m
Therefore, the combination rises approximately 1.275 meters above its initial position.
To solve this problem, we can use the principles of conservation of momentum and conservation of mechanical energy.
(a) Let's start by finding the speed of the bullet/block combination immediately after the collision.
1. Find the momentum of the bullet before the collision:
Momentum = mass * velocity
Bullet momentum = 2.86 g * 452 m/s
However, it's important to convert the mass from grams to kilograms (1 g = 0.001 kg):
Bullet momentum = 0.00286 kg * 452 m/s
2. Assume the bullet embeds itself into the wooden block, and both move as one afterward. Conservation of momentum tells us that the total momentum before the collision should equal the total momentum after the collision:
Total momentum before collision = Total momentum after collision
Bullet momentum before collision = (Bullet/Block combination mass) * (Bullet/Block combination velocity)
We need to find the velocity of the combination.
3. Find the total mass of the bullet/block combination:
Bullet/Block combination mass = Bullet mass + Block mass
Convert the block mass from grams to kilograms:
Block mass = 230 g * 0.001 kg/g
4. Substitute the values into the equation and solve for the velocity of the combination:
Bullet momentum before collision = (Bullet/Block combination mass) * (Bullet/Block combination velocity)
0.00286 kg * 452 m/s = (Bullet/Block combination mass) * (Bullet/Block combination velocity)
Simplifying the equation:
(Bullet/Block combination velocity) = (0.00286 kg * 452 m/s) / (Bullet/Block combination mass)
Calculate the value using the given masses and convert it to m/s.
(b) To find how high the combination rises above its initial position, we can use the conservation of mechanical energy.
1. Find the initial mechanical energy of the system, which is equal to the kinetic energy of the bullet before the collision:
Initial kinetic energy = (1/2) * mass * velocity^2
Initial kinetic energy = (1/2) * (2.86 g * 0.001 kg/g) * (452 m/s)^2
2. Find the final mechanical energy of the system when the bullet and block reach their maximum height. At maximum height, all of the initial kinetic energy is converted to potential energy due to gravity:
Final potential energy = mass * gravity * height
Since the block is at its maximum height and has gravitational potential energy, the combination of the bullet and block also has the same potential energy.
3. Set the initial kinetic energy equal to the final potential energy and solve for the height:
(1/2) * (2.86 g * 0.001 kg/g) * (452 m/s)^2 = (Bullet/Block combination mass) * gravity * height
Simplify the equation and solve for the height.
These calculations will provide you with the speed of the bullet/block combination immediately after the collision (part a) and the height the combination rises above its initial position (part b).