a 3.30 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second, with mass 1.80 kg. Speeds of 0.630 m/s and 1.40 m/s, respectively, are thereby given to the blocks (Fig. 7-28b). Neglect the mass removed from the first block by the bullet. Figure 7-28 (a) Find the speed of the bullet immediately after it emerges from the first block. m/s
Question
a 3.30 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second, with mass 1.80 kg. Speeds of 0.630 m/s and 1.40 m/s, respectively, are thereby given to the blocks (Fig. 7-28b). Neglect the mass removed from the first block by the bullet.
Figure 7-28 (a) Find the speed of the bullet immediately after it emerges from the first block. m/s
Solution
To solve this problem, we need to use the principle of conservation of momentum. The total momentum before the bullet hits the first block is equal to the total momentum after it passes through the first block.
Before the bullet hits the first block, the total momentum is just the momentum of the bullet, which is the mass of the bullet times its initial speed. We don't know the initial speed of the bullet, so let's call it v. The momentum is then 3.30 g * v.
After the bullet passes through the first block, the total momentum is the sum of the momentum of the bullet, the first block, and the second block. The bullet and the first block are moving together at a speed of 0.630 m/s, so their combined momentum is (3.30 g + 1.20 kg) * 0.630 m/s. The second block is still at rest, so its momentum is 0.
Setting these two expressions for the total momentum equal to each other gives us the equation:
3.30 g * v = (3.30 g + 1.20 kg) * 0.630 m/s
Solving this equation for v will give us the speed of the bullet immediately after it emerges from the first block.
Note: Be careful with units. The mass of the bullet is given in grams, but the masses of the blocks are given in kilograms. You'll need to convert the mass of the bullet to kilograms before you can solve the equation. 1 g = 0.001 kg, so the mass of the bullet is 0.0033 kg.
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