A 5.00 g bullet moving at 648 m/s strikes a 670 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 440 m/s.

A 5.00 g bullet moving at 648 m/s strikes a 670 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 440 m/s.
(a) What is the resulting speed of the block?

(b) What is the speed of the bullet-block center of mass?
(b) What is the speed of the bullet-block center of mass?

Momentum before = 648 * .005

equals
Momentum after = 440*.005 + v * .670
solve for v
v = 1.55 m/s

Speed of center of mass does not change because total mass does not change

b)before : average speed = (648*.005 + 0*.670)/.675 = 4.8 m/s

c) after: average speed = (440*.005 + 1.55*.670)/.675 = 4.8 m/s sure enough

Thank you :)

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To find the resulting speed of the block after the bullet emerges, we need to use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's first calculate the momentum before the collision. The momentum of an object is given by the product of its mass and velocity. The bullet has a mass of 5.00 g, which is equal to 0.005 kg, and a velocity of 648 m/s. The block is at rest, so its velocity is 0.

Momentum of bullet before collision = mass * velocity = 0.005 kg * 648 m/s

To find the momentum after the collision, we can calculate the momentum of the bullet using its new velocity of 440 m/s. The block will have a velocity after the collision, which we need to find.

Momentum of bullet after collision = mass * velocity = 0.005 kg * 440 m/s

Since the total momentum before the collision is equal to the total momentum after the collision, we can write an equation:

Momentum before = Momentum after

0.005 kg * 648 m/s = 0.005 kg * 440 m/s + 670 g * velocity of the block

Now, let's solve for the velocity of the block:

0.005 kg * 648 m/s = 0.005 kg * 440 m/s + 0.670 kg * velocity of the block

Simplifying the equation:

(0.005 kg * 648 m/s) - (0.005 kg * 440 m/s) = 0.670 kg * velocity of the block

0.810 kg·m/s = 0.670 kg * velocity of the block

Dividing both sides by 0.670 kg:

velocity of the block = 0.810 kg·m/s / 0.670 kg

Therefore, the resulting speed of the block is approximately 1.208 m/s.

To find the speed of the bullet-block center of mass, we need to use the concept of the conservation of linear momentum. The center of mass velocity is equal to the total momentum divided by the total mass.

The total momentum before the collision is the momentum of the bullet, which is 0.005 kg * 648 m/s, and the total mass is the sum of the bullet and the block masses, which is 0.005 kg + 0.670 kg.

Using these values, we can calculate the speed of the bullet-block center of mass:

Speed of bullet-block center of mass = Total momentum / Total mass

Speed of bullet-block center of mass = (0.005 kg * 648 m/s) / (0.005 kg + 0.670 kg)

After performing the calculation, we find that the speed of the bullet-block center of mass is approximately 647.654 m/s.