A block of wood of mass 0.86 kg is initially at rest. A bullet of mass 0.048 kg traveling at 102.3 m/s strikes the block and becomes embedded in it. With what speed do the block of wood and the bullet move just after the collision?
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To determine the speed at which the block of wood and the bullet move just after the collision, we can apply the principles of conservation of momentum.
First, let's calculate the initial momentum of the bullet before the collision and the block of wood. The momentum (p) is given by the equation:
p = mass × velocity
Initial momentum of the bullet = 0.048 kg × 102.3 m/s = 4.9184 kg·m/s
Since the block of wood is initially at rest, its initial momentum is zero.
According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision. Therefore:
Initial momentum of the bullet + Initial momentum of the block = Final momentum of the bullet + Final momentum of the block
0.048 kg × 102.3 m/s + 0 kg × 0 m/s = (0.048 kg + 0.86 kg) × v
4.9184 kg·m/s = 0.908 kg × v
Now, we can solve for the final velocity (v) by dividing both sides of the equation by 0.908 kg:
v = 4.9184 kg·m/s / 0.908 kg
v ≈ 5.41 m/s
Therefore, the block of wood and the bullet move with a speed of approximately 5.41 m/s just after the collision.