In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.13 kg) and embeds itself in block 2 (mass 1.75 kg). The blocks end up with speeds v1 = 0.600 m/s and v2 = 1.32 m/s (see Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) enters and (b) leaves block 1.
Well, I have to say, this sounds like a pretty bullet-proof situation! Let's break it down.
(a) When the bullet enters block 1, it's like entering a bad relationship - it's going to slow down. But how much? To find out, we need to use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
We know that the bullet has a mass of 3.50 g, which is 0.0035 kg. The mass of block 1 is 1.13 kg. So, we can write the equation:
(m_bullet * v_bullet_before) + (m_block1 * v_block1_before) = (m_bullet * v_bullet_after) + (m_block1 * v_block1_after)
Plugging in the values we know, we get:
(0.0035 kg * v_bullet_before) + (1.13 kg * 0 m/s) = (0.0035 kg * v_bullet_after) + (1.13 kg * 0.600 m/s)
Solving for v_bullet_before, we find:
v_bullet_before = (0.0035 kg * v_bullet_after) / 0.0035 kg
And voila! You have the speed of the bullet as it enters block 1.
(b) Now, when the bullet leaves block 1, it's like getting out of a bad relationship - it wants to pick up speed again! So, we use the same principle of conservation of momentum to figure out how fast it's going.
Now, the mass of block 2 is 1.75 kg, and we know its speed after the collision is 1.32 m/s. So, the equation becomes:
(0.0035 kg * v_bullet_after) + (1.13 kg * 0.600 m/s) = (0.0035 kg * v_bullet_after) + (1.75 kg * v_block2_after)
Simplifying this, we get:
(0.0035 kg * v_bullet_after) = (1.75 kg * v_block2_after) - (1.13 kg * 0.600 m/s)
And there you have it! You can now calculate the speed of the bullet as it leaves block 1. Just be sure to watch out for any emotional baggage it might carry!
To solve this problem, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Step 1: Find the initial momentum of the bullet before it hits block 1.
The initial momentum of an object is given by the equation:
p_initial = m_bullet * v_initial
where m_bullet is the mass of the bullet and v_initial is the initial velocity of the bullet.
Step 2: Find the final momentum of the system after the collision.
The final momentum of the system is the sum of the momenta of block 1 and block 2.
p_final = (m_block1 * v1) + (m_block2 * v2)
where m_block1 is the mass of block 1, v1 is its final velocity, m_block2 is the mass of block 2, and v2 is its final velocity.
Step 3: Determine the change in momentum of the bullet.
The change in momentum of the bullet is equal to the final momentum minus the initial momentum.
Δp = p_final - p_initial
Step 4: Apply the law of conservation of momentum.
Since the total momentum before the collision is equal to the total momentum after the collision, we can set the initial momentum of the bullet equal to the change in momentum:
p_initial = Δp
Step 5: Solve for the initial and final velocities of the bullet.
Knowing the mass of the bullet and the change in momentum, we can use the equation:
v_initial = Δp / m_bullet
v_final = v1
Let's calculate the values:
Step 1: Initial momentum of the bullet
p_initial = (3.50 g) * v_initial
Step 2: Final momentum of the system
p_final = (1.13 kg) * v1 + (1.75 kg) * v2
Step 3: Change in momentum of the bullet
Δp = p_final - p_initial
Step 4: Apply the law of conservation of momentum
p_initial = Δp
Step 5: Solve for the initial and final velocities of the bullet
v_initial = Δp / (3.50 g)
v_final = v1
So, the speed of the bullet as it enters block 1 (v_initial) is Δp / (3.50 g) and the speed of the bullet as it leaves block 1 (v_final) is v1.
To find the speed of the bullet as it enters and leaves block 1, we can apply the principle of conservation of linear momentum.
According to the principle of conservation of linear momentum, the total momentum of an isolated system remains constant before and after a collision. In this case, the isolated system consists of the bullet and the two blocks.
Let's denote the initial velocity of the bullet as u and its final velocity as v. We need to find both u and v.
Before the collision, the bullet has only horizontal motion. Therefore, its momentum is given by:
P_initial = m_bullet * u,
where m_bullet is the mass of the bullet (3.50 g = 0.00350 kg).
After the collision, the bullet embeds itself in block 2. The momentum of the combined system of block 2 and the bullet is:
P_final = (m_block2 + m_bullet) * v2,
where m_block2 is the mass of block 2 (1.75 kg) and v2 is the final velocity of block 2 (1.32 m/s).
Similarly, the momentum of block 1 after the collision is:
P_block1 = m_block1 * v1,
where m_block1 is the mass of block 1 (1.13 kg) and v1 is the final velocity of block 1 (0.600 m/s).
Using the conservation of linear momentum, we can equate the total initial momentum with the total final momentum:
P_initial = P_final + P_block1.
Substituting the respective momentum expressions, we have:
m_bullet * u = (m_block2 + m_bullet) * v2 + m_block1 * v1.
To solve for u, we isolate it in the equation:
u = [(m_block2 + m_bullet) * v2 + m_block1 * v1] / m_bullet.
Now we have the speed of the bullet as it enters block 1 (u). To find the speed of the bullet as it leaves block 1 (v), we need to consider the momentum just after the bullet embeds in block 2.
The momentum of the combined system of block 2 and the bullet just after the collision is:
P_final_leave = (m_block2 + m_bullet) * v,
where v is the final velocity of the combined system.
Again, using the conservation of linear momentum, we equate it to the total initial momentum:
P_initial = P_final_leave + P_block1.
Substituting the respective momentum expressions, we have:
m_bullet * u = (m_block2 + m_bullet) * v + m_block1 * v1.
To solve for v, we isolate it in the equation:
v = (m_bullet * u - m_block1 * v1) / (m_block2 + m_bullet).
Now we have the speed of the bullet as it enters block 1 (u) and as it leaves block 1 (v).
I hope this explanation helps you understand how to approach and solve this problem using the principle of conservation of linear momentum.