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Advanced Physics Homework Help
Calculate Speed of 2kg Block After Clay Hits and Sticks
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[QUOTE="NateD, post: 6886822, member: 133374"] To solve this problem, we need to use the conservation of momentum. According to this principle, the total momentum of the system (the block and clay ball) before the collision must be equal to the total momentum of the system after the collision. Before the collision, the momentum of the block is zero since it is initially at rest. The momentum of the clay ball is equal to its mass times its velocity, mv, so in this case the clay ball has a momentum of 0.1 kg x V, where V is the speed of the clay ball. Therefore, the total momentum of the system before the collision is 0.1V.After the collision, the momentum of the combined block-clay ball system is still 0.1V, since the momentum of the block is still zero. The kinetic energy of this system is equal to the sum of the kinetic energies of the block and clay ball. The kinetic energy of the block is equal to 1/2mv^2, while the kinetic energy of the clay ball is equal to 1/2(0.1)v^2. Thus, the total kinetic energy of the combined system is equal to 1/2(2.1)v^2. Setting the total momentum of the system before the collision equal to the total kinetic energy of the system after the collision, we get:0.1V = 1/2(2.1)v^2 Solving for v, we get:V = (2.1)v^2 / 0.1 v = sqrt(V*0.1/2.1) Therefore, the speed of the clay ball and 2kg block immediately after the clay sticks to the block but before the spring compresses significantly is equal to the square root of V*0.1/2.1. [/QUOTE]
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Calculate Speed of 2kg Block After Clay Hits and Sticks
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