A 15.0 kgkg block is attached to a very light horizontal spring of force constant 375 N/mN/m and is resting on a smooth horizontal table (Figure 1). Suddenly it is struck by a 3.00 kgkg stone traveling horizontally at 8.00 m/sm/s to the right, whereupon the stone rebounds at 2.00 m/sm/s horizontally to the left.Figure1 of 1Part AFind the maximum distance that the block will compress the spring after the collision. (Hint: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)Express your answer in meters.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typefor Part AActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typefor Part Aundo for Part Aredo for Part Areset for Part Akeyboard shortcuts for Part Ahelp for Part Ax𝑥 =
Question
A 15.0 kgkg block is attached to a very light horizontal spring of force constant 375 N/mN/m and is resting on a smooth horizontal table (Figure 1). Suddenly it is struck by a 3.00 kgkg stone traveling horizontally at 8.00 m/sm/s to the right, whereupon the stone rebounds at 2.00 m/sm/s horizontally to the left.Figure1 of 1Part AFind the maximum distance that the block will compress the spring after the collision. (Hint: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)Express your answer in meters.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typefor Part AActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typefor Part Aundo for Part Aredo for Part Areset for Part Akeyboard shortcuts for Part Ahelp for Part Ax𝑥 =
Solution
To solve this problem, we need to break it down into two parts: the collision and the behavior after the collision.
Part 1: The Collision We can use the principle of conservation of momentum for this part. The total momentum before the collision is equal to the total momentum after the collision.
The initial momentum of the system is the momentum of the stone, as the block is initially at rest. The momentum of an object is given by the product of its mass and velocity. So, the initial momentum (P_initial) is:
P_initial = mass_stone * velocity_stone = 3.00 kg * 8.00 m/s = 24.0 kg*m/s
The final momentum of the system is the sum of the momentum of the stone and the block after the collision. The stone rebounds with a velocity of 2.00 m/s to the left, which we'll take as negative. So, the final momentum (P_final) is:
P_final = mass_stone * velocity_stone + mass_block * velocity_block
We can set the initial momentum equal to the final momentum and solve for the velocity of the block (velocity_block):
24.0 kgm/s = -3.00 kg * 2.00 m/s + 15.0 kg * velocity_block 24.0 kgm/s = -6.00 kgm/s + 15.0 kg * velocity_block 30.0 kgm/s = 15.0 kg * velocity_block velocity_block = 30.0 kg*m/s / 15.0 kg = 2.00 m/s
Part 2: The Behavior After the Collision After the collision, the block compresses the spring and comes to rest momentarily when the spring is fully compressed. We can use the principle of conservation of energy for this part. The initial kinetic energy of the block is converted into potential energy in the spring at maximum compression.
The kinetic energy (KE) of an object is given by 1/2 * mass * velocity^2. So, the initial kinetic energy (KE_initial) of the block is:
KE_initial = 1/2 * mass_block * velocity_block^2 = 1/2 * 15.0 kg * (2.00 m/s)^2 = 30.0 J
The potential energy (PE) stored in a spring is given by 1/2 * k * x^2, where k is the spring constant and x is the displacement (compression in this case). At maximum compression, all the kinetic energy is converted into potential energy. So, we can set the initial kinetic energy equal to the potential energy and solve for the displacement (x):
30.0 J = 1/2 * 375 N/m * x^2 60.0 J = 375 N/m * x^2 x^2 = 60.0 J / 375 N/m = 0.16 m^2 x = sqrt(0.16 m^2) = 0.40 m
So, the maximum distance that the block will compress the spring after the collision is 0.40 meters.
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