Question

3. A rock is dropped (from rest) from the top of a 60 m tall building. How far above the ground is the rock 1.2 s before it reaches the ground?
4. A mother racing her son has half the kinetic energy of the son, who has half the mass as the mother. Mother speeds up by 1.0 m/s and then has the same kinetic energy as that of the son. What are the original speeds of the mother and the son?

Answer 1

Answer on Question #42194, Physics, Mechanics | Kinematics | Dynamics

3. A rock is dropped (from rest) from the top of a 60 m tall building. How far above the ground is the rock 1.2 s before it reaches the ground?

4. A mother racing her son has half the kinetic energy of the son, who has half the mass as the mother. Mother speeds up by 1.0 m/s and then has the same kinetic energy as that of the son. What are the original speeds of the mother and the son?

Solution:

3.

Given:

$$y_0 = 60 \, \text{m},$$ $$t = 1.2 \, \text{s},$$ $$y_1 = ?$$

An object in free fall experiences an acceleration $g$ of $-9.81 \, \text{m/s}^2$. (The sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is $-9.81 \, \text{m/s}^2$ for any freely falling object.

The equation of moving to the top point is

$$y = y_0 + v_0 t + \frac{1}{2} a t^2$$

where

$$y_0 = 60 \, \text{m} \text{ is initial position}$$ $$v_0 = 0 \text{ is initial speed}$$ $$a = g = -9.81 \, \text{m/s}^2 \text{ is acceleration}$$

At time $t = 1.2 \, \text{s}$, the position of a rock is

$$y_1 = 60 + 0 + \frac{1}{2}(-9.81) \cdot 1.2^2 = 52.94 \, \text{m}$$

Answer. $y_1 = 52.94 \, \text{m}$.

4. A mother racing her son has half the kinetic energy of the son, who has half the mass as the mother. Mother speeds up by $1.0 \, \text{m/s}$ and then has the same kinetic energy as that of the son. What are the original speeds of the mother and the son?

Solution:

Given:

$$\begin{array}{l} m_{2} = m_{1}/2, \\ KE_{1} = KE_{2}/2 \\ v_{1} = ?, \\ v_{2} = ? \end{array}$$

The kinetic energy is

$$\begin{array}{l} KE_{1} = \frac{KE_{2}}{2} \\ \frac{m_{1}v_{1}^{2}}{2} = \frac{m_{2}v_{2}^{2}}{4} \end{array}$$

So,

$$\begin{array}{l} 2m_{1}v_{1}^{2} = m_{2}v_{2}^{2} \\ m_{2} = m_{1}/2 \end{array}$$

Thus,

$$2m_{1}v_{1}^{2} = \frac{m_{1}}{2}v_{2}^{2}$$ $$4v_{1}^{2} = v_{2}^{2}$$

From given we also have

$$\begin{array}{l} \frac{m_{1}(v_{1} + 1)^{2}}{2} = \frac{m_{2}v_{2}^{2}}{2} \\ \frac{m_{1}(v_{1} + 1)^{2}}{2} = \frac{m_{1}v_{2}^{2}}{4} \end{array}$$

Thus, the second equation is

$$2(v_{1} + 1)^{2} = v_{2}^{2}$$ $$\begin{array}{l} 2(v_{1} + 1)^{2} = 4v_{1}^{2} \\ 2(v_{1}^{2} + 2v_{1} + 1) = 4v_{1}^{2} \end{array}$$

We obtain quadratic equation

$$2v_{1}^{2} - 2v_{1} - 1 = 0$$

Solution is

$$\begin{array}{l} v_{1} = \frac{2 + \sqrt{4 + 4 \cdot 2}}{4} = 1.366 \, \text{m/s} \\ v_{2} = 2v_{1} = 2 \cdot 1.366 = 2.732 \, \text{m/s} \end{array}$$

Answer. The mother speed is $v_{1} = 1.366 \, \text{m/s}$,

the son speed is $v_{2} = 2.732 \, \text{m/s}$

http://www.AssignmentExpert.com/