A 35.0-g bullet moving at 475 m/s strikes a 2.5-kg wooden block. The bullet passes through the block, leaving at 275 m/s. The block was at rest when it was hit. How fast is it moving when the bullet leaves?
.035 * 475 = initial momentum
.035*275 + 2.5 * v = final momentum
initial momentum = final momentum. Solve for v
Well, well, well! Looks like we've got ourselves a speedy bullet and a block that's about to experience some high-speed action! Now let me do some clown calculations for you.
First things first, let's find out how much momentum our bullet initially has. Momentum is mass times velocity (p = mv). So, the momentum of the bullet is 35.0 grams (or 0.035 kg) multiplied by 475 m/s. Can you guess the answer? It's a real momentum-tous question!
The momentum of the bullet initially is 16.625 kg*m/s. Impressive, isn't it?
Now, when the bullet passes through the block, their total momentum should stay the same. That means the block should gain momentum equal to the initial momentum of the bullet.
Since the block's mass is 2.5 kg and it was initially at rest, the final velocity of the block can be found using the equation p = mv. So, 16.625 kg*m/s = 2.5 kg * v.
Dividing both sides by 2.5 kg, we find that the velocity of the block is approximately 6.65 m/s, or about the speed at which I chase after my dreams!
So, when the bullet leaves the block, it'll be moving at a speed of around 6.65 m/s. Keep on rolling with those physics questions, my friend!
To solve this problem, we can use the principle of conservation of linear momentum. The total momentum before the collision is equal to the total momentum after the collision.
The momentum of the bullet before the collision is given by:
p1 = m1 * v1
where,
m1 = mass of the bullet = 35.0 g = 0.035 kg
v1 = velocity of the bullet before the collision = 475 m/s
The momentum of the block before the collision is given by:
p2 = m2 * v2
where,
m2 = mass of the block = 2.5 kg
v2 = velocity of the block before the collision (which is at rest) = 0 m/s
After the collision, the bullet passes through the block, so the total momentum is given by:
p3 = m1 * v3
where,
v3 = velocity of the bullet after the collision = 275 m/s
The block gains the momentum of the bullet, so its momentum after the collision is given by:
p4 = m2 * v4
where,
v4 = velocity of the block after the collision (which we need to find)
According to the principle of conservation of linear momentum, we have:
p1 + p2 = p3 + p4
Substituting the values, we get:
(m1 * v1) + (m2 * v2) = (m1 * v3) + (m2 * v4)
(0.035 kg * 475 m/s) + (2.5 kg * 0 m/s) = (0.035 kg * 275 m/s) + (2.5 kg * v4)
Simplifying:
0.035 * 475 = 0.035 * 275 + 2.5 * v4
16.625 = 9.625 + 2.5 * v4
Now, we can solve for v4:
2.5 * v4 = 16.625 - 9.625
2.5 * v4 = 7
v4 = 7 / 2.5
v4 = 2.8 m/s
Therefore, the block is moving at a speed of 2.8 m/s when the bullet leaves.
To find the speed of the wooden block after the bullet passes through it, we can use the principle of conservation of momentum. According to this principle, the initial momentum of the system (bullet + block) should be equal to the final momentum of the system.
The momentum of an object is calculated by multiplying its mass by its velocity. So, the initial momentum of the system can be found by adding the momentum of the bullet and the momentum of the block before the collision:
Initial momentum = (mass of bullet × velocity of bullet) + (mass of block × velocity of block)
Given:
Mass of bullet (m1) = 35.0 g = 0.035 kg
Velocity of bullet (v1) = 475 m/s
Mass of block (m2) = 2.5 kg
Velocity of block (v2) = 0 m/s (since the block was at rest)
Initial momentum = (0.035 kg × 475 m/s) + (2.5 kg × 0 m/s)
= 1.225 kg⋅m/s + 0 kg⋅m/s
= 1.225 kg⋅m/s
Now, let's calculate the final momentum of the system after the bullet passes through the block. We can again use the principle of conservation of momentum:
Final momentum = (mass of bullet × velocity of bullet after passing through) + (mass of block × velocity of block after passing through)
Given:
Velocity of bullet after passing through (v3) = 275 m/s
Final momentum = (0.035 kg × 275 m/s) + (2.5 kg × v4)
Since we're trying to find the velocity of the block after the bullet passes through, let's represent that as v4.
Now, we set up an equation by equating the initial momentum and final momentum:
1.225 kg⋅m/s = (0.035 kg × 275 m/s) + (2.5 kg × v4)
We can now solve this equation to find the velocity of the block (v4).
1.225 kg⋅m/s = 9.625 kg⋅m/s + (2.5 kg × v4)
1.225 kg⋅m/s - 9.625 kg⋅m/s = 2.5 kg × v4
-8.4 kg⋅m/s = 2.5 kg × v4
v4 = (-8.4 kg⋅m/s) / 2.5 kg
v4 = -3.36 m/s
The negative sign indicates that the block is moving in the opposite direction of the bullet. Therefore, the magnitude of the velocity of the block after the bullet passes through is 3.36 m/s.