A bullet of mass 4.0 g is fired horizontally into a 2.4 kg wooden block at rest on a horizontal surface. The bullet stops in the block. The speed of the block immediately after the bullet stops relative to it is 2.4 m/s. At what speed is the bullet fired?
Question
A bullet of mass 4.0 g is fired horizontally into a 2.4 kg wooden block at rest on a horizontal surface. The bullet stops in the block. The speed of the block immediately after the bullet stops relative to it is 2.4 m/s. At what speed is the bullet fired?
Solution
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision (bullet fired into the block) is equal to the total momentum after the collision (bullet stopped in the block).
The formula for momentum is mass times velocity (p=mv).
Before the collision, the momentum is only from the bullet because the block is at rest. Let's denote the bullet's speed as v. So, the momentum before the collision is (mass of bullet) * (speed of bullet) = 0.004 kg * v.
After the collision, the bullet is stopped within the block and they move together. So, the momentum after the collision is (mass of bullet + mass of block) * (speed of block) = (0.004 kg + 2.4 kg) * 2.4 m/s.
According to the conservation of momentum, these two should be equal:
0.004 kg * v = (0.004 kg + 2.4 kg) * 2.4 m/s.
Solving this equation for v gives us the speed at which the bullet was fired.
v = (0.004 kg + 2.4 kg) * 2.4 m/s / 0.004 kg = 1440 m/s.
So, the bullet was fired at a speed of 1440 m/s.
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