A 3.5 g bullet is fired horizontally at 2 blocks resting on a smooth tabletop. The bullet passes trough the 1st block, with mass 1.2 kg and embed itself in the second with mass 1.8 kg. Speeds of 0.63 m/s and 1.4 m/s, respectively are there by imparted on the blocks. Neglecting the mass removed from the first block by the bullet, find (a) the speed of the bullet immediately after it emerges from the 1st block and (b) bullets original speed.
To solve this problem, we can make use of the principle of conservation of momentum.
(a) The speed of the bullet immediately after it emerges from the 1st block:
First, we need to determine the total momentum before the bullet emerges from the first block. We know that the first block has a mass of 1.2 kg and a speed of 0.63 m/s, while the bullet has a mass of 3.5 g (0.0035 kg) and an unknown speed v1.
The momentum before emerging from the first block is given by:
Momentum before = (mass of 1st block * velocity of 1st block) + (mass of bullet * velocity of bullet)
Pbefore = (1.2 kg * 0.63 m/s) + (0.0035 kg * v1)
Next, we can determine the momentum after the bullet emerges from the first block. In this case, the first block no longer contributes to the momentum, and the bullet has a mass of 3.5 g (0.0035 kg) and a speed of v2 after emerging.
The momentum after = (mass of bullet * velocity of bullet after emerging)
Pafter = (0.0035 kg * v2)
According to the principle of conservation of momentum, the total momentum before emerging is equal to the total momentum after emerging:
Pbefore = Pafter
Therefore, we can set the two expressions equal:
(1.2 kg * 0.63 m/s) + (0.0035 kg * v1) = (0.0035 kg * v2)
Simplifying the equation:
0.756 kg•m/s + 0.0035 kg•m/s•v1 = 0.0035 kg•m/s•v2
We are given that the bullet embeds itself in the second block, so its velocity after emerging from the first block will be the same as the velocity of the second block. Thus, we can substitute v2 with the given value of 1.4 m/s:
0.756 kg•m/s + 0.0035 kg•m/s•v1 = 0.0035 kg•m/s•1.4 m/s
Now we can solve for v1:
0.756 kg•m/s + 0.0035 kg•m/s•v1 = 0.0049 kg•m/s
0.0035 kg•m/s•v1 = 0.0049 kg•m/s - 0.756 kg•m/s
0.0035 kg•m/s•v1 = 0.00414 kg•m/s
v1 = 0.00414 kg•m/s / 0.0035 kg
v1 = 1.18286 m/s
Therefore, the speed of the bullet immediately after it emerges from the first block is approximately 1.18 m/s.
(b) The bullet's original speed:
Since the bullet embeds itself in the second block, the speed of the second block after the collision is given as 1.4 m/s. We can use this information to calculate the original speed of the bullet.
The momentum before the collision is the sum of the momentum of the first block and the bullet:
Momentum before = (mass of 1st block * velocity of 1st block) + (mass of bullet * velocity of bullet)
Using the given values, let's substitute the variables:
(1.2 kg * 0.63 m/s) + (0.0035 kg * bullet's original speed) = (1.8 kg * 1.4 m/s)
Simplifying the equation:
0.756 kg•m/s + 0.0035 kg•m/s•bullet's original speed = 2.52 kg•m/s
0.0035 kg•m/s•bullet's original speed = 2.52 kg•m/s - 0.756 kg•m/s
0.0035 kg•m/s•bullet's original speed = 1.764 kg•m/s
bullet's original speed = 1.764 kg•m/s / 0.0035 kg
bullet's original speed = 504 m/s
Therefore, the bullet's original speed is approximately 504 m/s.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, provided there are no external forces acting on the system.
Let's first find the velocity of the 1st block after the bullet embeds itself into it. We can use the formula:
m1 * v1 = (m1 + m) * v
where m1 is the mass of the 1st block (1.2 kg), v1 is the initial velocity of the 1st block, m is the mass of the bullet (3.5 g = 0.0035 kg), and v is the final velocity of the combined mass (v1 = 0 because the block starts at rest).
Plugging in the values, we have:
1.2 kg * 0 = (1.2 kg + 0.0035 kg) * v
Simplifying the equation:
0 = 1.2035 kg * v
v = 0
Therefore, the 1st block comes to rest after the bullet embeds itself into it.
Now, let's find the velocity of the 2nd block after being hit by the bullet. We can use the same conservation of momentum principle:
(m1 + m) * v = m2 * v2
where m2 is the mass of the 2nd block (1.8 kg), v2 is the final velocity of the 2nd block, m is the mass of the bullet (0.0035 kg), and v is the final velocity of the combined mass (v2 = 0 because the block is at rest initially).
Plugging in the values, we have:
(1.2 kg + 0.0035 kg) * v = 1.8 kg * 1.4 m/s
Simplifying the equation:
1.2035 kg * v = 2.52 kg*m/s
v = 2.52 kg*m/s / 1.2035 kg
v ≈ 2.096 m/s
Therefore, the velocity of the 2nd block after being hit by the bullet is approximately 2.096 m/s.
Now, let's find the speed of the bullet immediately after it emerges from the 1st block. Since the bullet is embedded into the 1st block and both are at rest, the velocity of the bullet right after emerging is equal to the velocity of the 1st block, which is 0.
Therefore, the speed of the bullet immediately after it emerges from the 1st block is 0 m/s.
Finally, let's find the original speed of the bullet. We can use the conservation of momentum again to solve for this.
m1 * v1 = (m1 + m) * v
1.2 kg * v1 = (1.2 kg + 0.0035 kg) * 2.096 m/s
Simplifying the equation:
1.2 kg * v1 = 1.2035 kg * 2.096 m/s
v1 = (1.2035 kg * 2.096 m/s) / 1.2 kg
v1 ≈ 2.097 m/s
Therefore, the original speed of the bullet is approximately 2.097 m/s.
about the second block first:
m v before = m v after
0.0035 v = 1.80035 (1.4)
v = 720 m/s
well that was part (a)
now part (b)
0.0035 u = 1.2 (0.63) + 0.0035 (720)
solve for u