A bullet with a mass of 5.40 g is fired horizontally at two blocks resting on a smooth and frictionless table top as shown in the Figure. The bullet passes through the first 1.20 kg block, and embeds itself in a second 1.20 kg block. Speeds v1 = 3.90 m/s and v2 = 1.70 m/s, are thereby imparted on the blocks. The mass removed from the first block by the bullet can be neglected. Find the speed of the bullet immediately after emerging from the first block.

Question

A bullet with a mass of 5.40 g is fired horizontally at two blocks resting on a smooth and frictionless table top as shown in the Figure. The bullet passes through the first 1.20 kg block, and embeds itself in a second 1.20 kg block. Speeds v1 = 3.90 m/s and v2 = 1.70 m/s, are thereby imparted on the blocks. The mass removed from the first block by the bullet can be neglected. Find the speed of the bullet immediately after emerging from the first block.

Find the initial speed of the bullet.

Answer 1

Which block has speed v1, block 2?

Answer 2

I am not sure. I think the speeds are for the bullet before and after passing through the block. However, in the diagram the second block is in a different spot after the bullet hits it.

Answer 3

To find the initial speed of the bullet, we need to use the principle of conservation of linear momentum. According to this principle, the total momentum before the interaction is equal to the total momentum after the interaction.

The total momentum of the system before the interaction can be calculated by multiplying the mass of the bullet by its initial velocity, which we need to find. Let's call it v0.

Momentum before interaction = (Mass of Bullet) x (Initial Velocity of Bullet) = (5.40 g) x v0

After passing through the first block, the bullet embeds itself in the second block, so the combined momentum of the two blocks after the interaction can be calculated by adding the individual momenta of each block.

Momentum after interaction = (Mass of First Block) x (Final Velocity of First Block) + (Mass of Second Block) x (Final Velocity of Second Block)
= (1.20 kg) x (3.90 m/s) + (1.20 kg) x (1.70 m/s)
= 4.68 kg·m/s + 2.04 kg·m/s
= 6.72 kg·m/s

According to the principle of conservation of linear momentum, the total momentum before the interaction is equal to the total momentum after the interaction. Therefore, we can set up an equation equating the two:

(5.40 g) x v0 = 6.72 kg·m/s

To solve for v0, we first need to convert the mass of the bullet from grams to kilograms:

Mass of Bullet = 5.40 g = 0.0054 kg

Now we can solve for v0:

(0.0054 kg) x v0 = 6.72 kg·m/s
v0 = 6.72 kg·m/s / 0.0054 kg
v0 ≈ 1244.44 m/s

So the initial speed of the bullet is approximately 1244.44 m/s.