A 4.80 g bullet moving at 660 m/s strikes a 700 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 420 m/s. What is the resulting speed of the block? m/s
Question
A 4.80 g bullet moving at 660 m/s strikes a 700 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 420 m/s. What is the resulting speed of the block? m/s
Solution
This problem can be solved using the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Step 1: Calculate the initial momentum of the bullet and the block. The initial momentum of the bullet (Pb_initial) is its mass times its velocity, which is 4.80 g * 660 m/s = 3168 g*m/s. Since the block is initially at rest, its initial momentum (Pb_initial) is 0.
Step 2: Calculate the final momentum of the bullet. The final momentum of the bullet (Pb_final) is its mass times its final velocity, which is 4.80 g * 420 m/s = 2016 g*m/s.
Step 3: Calculate the final momentum of the block. According to the conservation of momentum, the final momentum of the block (Pb_final) is the initial total momentum minus the final momentum of the bullet, which is 3168 gm/s - 2016 gm/s = 1152 g*m/s.
Step 4: Calculate the final velocity of the block. The final velocity of the block (Vb_final) is its final momentum divided by its mass, which is 1152 g*m/s / 700 g = 1.65 m/s.
So, the resulting speed of the block is 1.65 m/s.
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