Question

a boy and his father run together.mass of the boy is half of his father ; kinetic energy of father is half of his son.if the father increases his velocity 1 m/s , then the kinetic energy becomes equal of each . what is their initial velocity?
if they run with their own kinetic energy on a road which is inclined and the road makes an angle of 30 degree with horizontal, how much distance will each of them complete ?

Answer 1

Answer on Question #51623, Physics, Mechanics | Kinematics | Dynamics

A boy and his father run together. Mass of the boy is half of his father; kinetic energy of father is half of his son. If the father increases his velocity $1 \, \text{m/s}$, then the kinetic energy becomes equal of each. What is their initial velocity?

If they run with their own kinetic energy on a road which is inclined and the road makes an angle of 30 degree with horizontal, how much distance will each of them complete?

Solution:

We start with the given data. Let the father's mass be $M$, so that the son's mass must be equal to $\frac{1}{2} M$. We put the father's velocity as $V_f$ and the son's velocity as $V_s$.

We know that the equation to find kinetic energy, $KE$, is the following, where $m$ is mass and $v$ is velocity:

$$KE = \frac{m v^2}{2}$$

Based on the above information we can apply this to our problem.

$$\frac{m_f V_f^2}{2} = \frac{1}{2} \left( \frac{m_f V_s^2}{2} \right)$$

From the noted equation we can express the value of $V_f^2$. Firstly we simplify the equation.

$$\frac{V_f^2}{2} = \frac{1}{2} \left( \frac{V_s^2}{2} \right)$$

We multiply both sides of the equation by 2 and obtained the following result.

$$V_f^2 = \frac{1}{2} V_s^2$$

Now we have to construct the equation which takes into account the following condition, the father increases his velocity $1 \, \text{m/s}$ and the kinetic energy becomes equal of each.

$$\frac{1}{2} \left( \frac{m_f (V_f + 1)^2}{2} \right) = \left( \frac{m_f V_f^2}{2} \right)$$

Simplify the equation by opening the parenthesis.

$$\frac{1}{2} V_f^2 = \frac{1}{4} \left( V_f^2 + 2 V_f + 1 \right)$$

Now we simplify by opening the parenthesis and combining like terms.

$$\frac{1}{2} V_f^2 - \frac{1}{4} V_f^2 - \frac{1}{2} V_f - \frac{1}{4} = 0$$

Then we need to solve the obtained quadratic equation for $V_f$.

$$\frac {1}{4} V _ {f} ^ {2} - \frac {1}{2} V _ {f} - \frac {1}{4} = 0$$

Multiply all terms by 4.

$$V _ {f} ^ {2} - 2 V _ {f} - 1 = 0$$ $$V _ {f _ {1, 2}} = \frac {- b \pm \sqrt {b ^ {2} - 4 a c}}{2 a}$$

We determine the first root.

$$V _ {f _ {1}} = \frac {2 + \sqrt {(2) ^ {2} - 4 (1) (- 1)}}{2 (1)} = \frac {2 + \sqrt {8}}{2} = 1 + \sqrt {2} = 2.4142 \mathrm{m/s}$$

We know that the second root will have the negative sign, thus, we accept the first solution.

$$V _ {f} = 2.4142 \mathrm{m/s}$$

Now we can calculate the son's velocity.

$$V _ {s} = 2 \cdot (2.4142) = 4.828 \mathrm{m/s}$$

The next part of the task is to determine the distance if we know their kinetic energy on a road which is inclined and the road makes an angle of 30 degree with horizontal.

Firstly we determine the father's kinetic energy.

$$\mathrm {K E} = \frac {\mathrm {m _ {f} V _ {f} ^ {2}}}{2} = \frac {\mathrm {m _ {f} (2 . 4 1 4) ^ {2}}}{2} = 2. 9 1 4 \mathrm{m _ {f}}$$

The distance to be held father will be equal

$$\begin{array}{l} \text {Distance} = \frac {\mathrm {K E}}{\mathrm {m _ {f} g s i n} \alpha} = \frac {\frac {\mathrm {m _ {f} V _ {f} ^ {2}}}{2}}{\mathrm {m _ {f} g s i n} \alpha} = \frac {\mathrm {m _ {f} V _ {f} ^ {2}}}{2 \mathrm {m _ {f} g s i n} \alpha} = \frac {\mathrm {V _ {f} ^ {2}}}{2 \mathrm {g s i n} \alpha} = \frac {(2 . 4 1 4) ^ {2}}{1 0} = \\ = \frac {5.827}{1 0} = 0.58 \mathrm{m} \\ \end{array}$$

Similarly we determine the distance for the son.

$$\text {Distance} = \frac {\mathrm {V _ {s} ^ {2}}}{2 \mathrm {g s i n} \alpha} = \frac {(4 . 8 2 8) ^ {2}}{1 0} = \frac {2 3 . 3 1 0}{1 0} = 2.3 \mathrm{m}$$

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