Mechanics Question

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A 2kg block is attached to an ideal spring (for which k = 200 N/m) and is initially at rest on a horizontal frictionless surface. A 100 gram ball of clay is thrown at the block. The clay is moving horizontally with speed V when it hits and sticks to the block. As a result, the pring is compressed a maximum distance of 0.4m.

a) Calculate the speed of the clay ball and 2kg block immediately after the clay sticks to the block but before the spring compresses significantly.


Ok, I kind of don't understand how the set-up is supposed to look like, so if someone can do ths problem, can they explain to me why they set it up the way they did?

THANKS :)
 

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  • #2
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.
 

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