Problem 26 - Master Resource Book in JEE Main Physics

Question

A man running has half the kinetic energy of a boy of half his mass. The man speeds up by \(1 \mathrm{~ms}^{-1}\) and then has kinetic energy as that of the boy. What were the original speeds of man and the boy? (a) \(\sqrt{2} \mathrm{~ms}^{-1} ; 2 \sqrt{2}-1 \mathrm{~ms}^{-1}\) (b) \((\sqrt{2}-1) \mathrm{ms}^{-1}, 2(\sqrt{2}-1) \mathrm{ms}^{-1}\) \((c)(\sqrt{2}+1) \mathrm{ms}^{-1} ; 2(\sqrt{2}+1) \mathrm{ms}^{-1}\) (d) None of the above

Step-by-Step Solution

Answer

Original speeds: man: \(\sqrt{2}+1\,\mathrm{ms}^{-1}\), boy: \(2(\sqrt{2}+1)\,\mathrm{ms}^{-1}\).

Step 1: Understand the given condition

We have two individuals: a man and a boy. The boy has half the mass of the man. Let the man's mass be denoted by \(m\) and his velocity by \(v_m\). Then, the boy's mass will be \(\frac{m}{2}\) and his velocity will be \(v_b\). We are told that the man's kinetic energy is initially half of the boy's kinetic energy.

Step 2: Write down the expressions for kinetic energies

The kinetic energy of the man is given by \( KE_{man} = \frac{1}{2} m v_m^2 \). The kinetic energy of the boy is given by \( KE_{boy} = \frac{1}{2} \left(\frac{m}{2}\right) v_b^2 \). According to the problem, \( \frac{1}{2} m v_m^2 = \frac{1}{2} * \frac{1}{2} \left(\frac{m}{2}\right) v_b^2 \). Simplifying, we get \( m v_m^2 = \frac{1}{2} * \frac{m}{2} v_b^2 \).

Step 3: Solve the energy equation for velocities

From the equation \( m v_m^2 = \frac{1}{2} \cdot \frac{m}{2} v_b^2 \), simplify to find \( v_m^2 = \frac{1}{4} v_b^2 \), hence \( v_m = \frac{1}{2} v_b \).

Step 4: Use the speed up condition

After speeding up by \( 1 \ \mathrm{ms}^{-1} \), the man's new velocity is \( v_m + 1 \). His new kinetic energy equals the boy's kinetic energy. \( \frac{1}{2} m (v_m + 1)^2 = \frac{1}{2} \cdot \frac{m}{2} v_b^2 \).

Step 5: Solve the new energy equation

Simplify the new kinetic energy equation: \( (v_m + 1)^2 = \frac{1}{2} v_b^2 \). Substitute \( v_m = \frac{1}{2}v_b \) into the equation: \( \left(\frac{v_b}{2} + 1\right)^2 = \frac{1}{2} v_b^2 \).

Step 6: Solve the quadratic equation

Expand and arrange \( \left(\frac{v_b}{2} + 1\right)^2 = \frac{1}{2} v_b^2 \) to \( \frac{v_b^2}{4} + v_b + 1 = \frac{1}{2} v_b^2 \). Rearranging gives you \( \frac{1}{2} v_b^2 - \frac{v_b^2}{4} - v_b - 1 = 0 \). Solving this quadratic equation yields \( v_b = 2(\sqrt{2} + 1) \). Hence, \( v_m = (\sqrt{2} + 1) \).

Key Concepts

Velocity Equations

When dealing with problems related to motion, understanding velocity is crucial. In this exercise, we have two people, a man and a boy, moving at different speeds. Velocity is a measure of how fast something is moving in a particular direction, measured in meters per second (m/s). In our case, the velocity of the man is denoted as \( v_m \) and the boy's velocity as \( v_b \). Initially, it is given that the man's kinetic energy, which depends on his velocity, is half that of the boy's. This means we can set up an equation to find out their velocities relative to each other. By using the relationship \( v_m = \frac{1}{2} v_b \), we find the man's initial velocity to be half of the boy's velocity. Knowing how to set up and solve these equations is essential in determining motion parameters in physics problems.

• Equation for man's velocity: \( v_m = \frac{1}{2} v_b \)
• Equation after man's speed increases: \( v_m + 1 \)

Quadratic Equations

Quadratic equations appear frequently in physics problems involving motion. These are polynomial equations of the form \( ax^2 + bx + c = 0 \). In our scenario, after the man speeds up, we required solving a quadratic equation to determine the boy's velocity. The equation was derived from equating the new kinetic energy of the man with that of the boy. Solving this required rearranging and expanding terms to obtain a clean quadratic form. You may need to use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the values of unknowns, depending on the complexity of the equation. In this exercise, solving the quadratic equation helped us find the boy's velocity in terms of the square root of expressions, which involves deeper mathematical skills.

• Identify the quadratic form: \( \frac{1}{2} v_b^2 - \frac{v_b^2}{4} - v_b - 1 = 0 \)
• Solve using the quadratic formula for \( v_b \)

Energy Conservation

Energy conservation is a pivotal concept when considering how kinetic energy changes in motion problems. This principle states that energy in a closed system remains constant. We see this principle at work in this problem when the man increases his speed by 1 m/s. His new kinetic energy becomes equal to that of the boy. Understanding kinetic energy, calculated as \( KE = \frac{1}{2} mv^2 \), is essential, as it varies with mass and velocity. In this exercise, initially, half of the man's kinetic energy matched the boy's full kinetic energy, illustrating how energy conservation permits prediction of subsequent states. By setting the condition after the man speeds up, we could equate the man's final kinetic energy with the boy’s existing energy and solve for the velocities using the quadratic equation outcomes. This demonstrates how energy conservation facilitates solving motion problems by linking initial and final states.

• Initial energy condition: man's energy is half the boy's.
• Final condition: man's new energy equals boy's energy.