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Chapter 7: Problem 55

A 6.0 -kg object is at rest on a perfectly frictionless surface when it is struck head-on by a 2.0 -kg object moving at \(10 \mathrm{m} / \mathrm{s} .\) If the collision is perfectly elastic, what is the speed of the 6.0-kg object after the collision? [Hint: You will need two equations.]

Short Answer

Expert verified
Answer: The speed of the 6.0-kg object after the perfectly elastic collision is \(\frac{5}{3} \mathrm{m} / \mathrm{s}\).

Step by step solution

01

Identify the variables

Before solving the problem, let's first identify the variables: - Mass of object 1 (m1) = 6.0 kg - Mass of object 2 (m2) = 2.0 kg - Initial velocity of object 1 (u1) = 0 m/s (at rest) - Initial velocity of object 2 (u2) = 10 m/s - Final velocity of object 1 (v1) = ? - Final velocity of object 2 (v2) Now that we've identified the variables, we can set up the equations for momentum and kinetic energy conservation.
02

Apply the conservation of momentum principle

The conservation of momentum principle states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be written as: \( m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2 \) Plugging in the values, we get: \( (6.0 \mathrm{kg})(0 \mathrm{m} / \mathrm{s}) + (2.0 \mathrm{kg})(10 \mathrm{m} / \mathrm{s}) = (6.0 \mathrm{kg})(v1) + (2.0 \mathrm{kg})(v2) \) Simplifying, we get: \( 20 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} = 6.0 \mathrm{kg} \cdot v1 + 2.0 \mathrm{kg} \cdot v2 \)(1)
03

Apply the conservation of kinetic energy principle

The conservation of kinetic energy principle states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Mathematically, this can be written as: \( \frac{1}{2}m1 * u1^2 + \frac{1}{2}m2 * u2^2 = \frac{1}{2}m1 * v1^2 + \frac{1}{2}m2 * v2^2 \) Plugging in the values, we get: \( \frac{1}{2}(6.0 \mathrm{kg})(0 \mathrm{m} / \mathrm{s})^2 + \frac{1}{2}(2.0 \mathrm{kg})(10 \mathrm{m} / \mathrm{s})^2 = \frac{1}{2}(6.0 \mathrm{kg})(v1)^2 + \frac{1}{2}(2.0 \mathrm{kg})(v2)^2 \) Simplifying, we get: \( 100 \mathrm{J} = 3.0 \mathrm{kg} \cdot v1^2 + 1.0 \mathrm{kg} \cdot v2^2 \) (2)
04

Solve the system of equations

Now we have a system of two equations with two unknowns (v1 and v2). To solve for v1, we can first solve equation (1) for v2 and then substitute this expression into equation (2): From equation (1), \( v2 = \frac{20 - 6.0v1}{2.0} \)(3) Substitute equation (3) into equation (2): \( 100 \mathrm{J} = 3.0 \mathrm{kg} \cdot v1^2 + 1.0 \mathrm{kg} \cdot \left(\frac{20 - 6.0v1}{2.0}\right)^2 \) Solve for v1: \( v1 = \frac{5}{3} \mathrm{m} / \mathrm{s} \)
05

State the final answer

The speed of the 6.0-kg object after the perfectly elastic collision is \(\frac{5}{3} \mathrm{m} / \mathrm{s}\).

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