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Grade 11Mechanics

A sphere of mass a frictionless horizontal table with velocity v it strikes with the spring force constant 1 min Newton per metre and compress 8 by 4 m the velocity of the sphere is

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8 Years agoGrade 11
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ApprovedApproved Tutor Answer1 Year ago

To solve the problem of a sphere colliding with a spring on a frictionless horizontal table, we can apply the principles of conservation of energy. When the sphere strikes the spring, its kinetic energy is converted into potential energy stored in the spring as it compresses. Let's break down the steps to find the velocity of the sphere after it compresses the spring.

Understanding the Energy Transformation

Initially, the sphere has kinetic energy due to its velocity. When it compresses the spring, this kinetic energy is transformed into elastic potential energy. The key formulas we will use are:

  • Kinetic Energy (KE) = (1/2)mv²
  • Potential Energy (PE) = (1/2)kx²

Where:

  • m = mass of the sphere
  • v = initial velocity of the sphere
  • k = spring constant (1 N/m in this case)
  • x = compression of the spring (0.4 m)

Setting Up the Equation

At the moment of maximum compression of the spring, all the kinetic energy of the sphere has been converted into potential energy. Therefore, we can set the kinetic energy equal to the potential energy:

(1/2)mv² = (1/2)kx²

Substituting Known Values

Given that:

  • k = 1 N/m
  • x = 0.4 m

We can substitute these values into our equation:

(1/2)mv² = (1/2)(1)(0.4)²

Now, simplifying the right side:

(1/2)mv² = (1/2)(1)(0.16) = 0.08

Solving for Velocity

Now, we can eliminate the (1/2) from both sides:

mv² = 0.16

To find the velocity, we rearrange the equation:

v² = 0.16/m

Taking the square root gives us:

v = √(0.16/m)

Final Thoughts

The final velocity of the sphere after compressing the spring depends on its mass. If you know the mass of the sphere, you can substitute that value into the equation to find the specific velocity. For example, if the mass were 0.5 kg, then:

v = √(0.16/0.5) = √(0.32) ≈ 0.566 m/s.

This approach illustrates how energy conservation principles can be applied to analyze the motion of objects interacting with springs. If you have any further questions or need clarification on any part of this process, feel free to ask!