In Problem 2, what is the speed of the car at (a) point A, (b) point B(c) point C?(d) How high will the car go on the last hill, which is too high for it to cross? (e) If we substitute a second car with twice the mass, what then are the answers to (a) through and (d)?

i) Mass of car$m=825\text{\hspace{0.17em}kg}$

ii) The height $h=42\text{\hspace{0.17em}m}$

iii) Gravitational acceleration $g=9.8{\text{m/s}}^{\text{2}}$

iv) Initial velocity$v_0=17\text{m/s}$

The mechanical energy $E_{mec}$of a system is the sum of its kinetic energy $K$and potential energy $U$.

Formula:

$K_2+U_2=K_1+U_1$

$K=\frac{1}{2}m{v}_0^2$

$U=mgh$

At point$A,U_A=mgh$ and at initial point$U_0=mgh.$

That means initial position of the car and point $A$ is on same height, so,

∴$U_A=U_0$

$K_0+U_0=K_A+U_A$

∴$K_0=K_A$

∴role="math" localid="1661168162281" $\begin{array}{c}{v}_0=v_A\\ =17.0\text{m/s}\end{array}$

The height of point$B$ from bottom$=h/2$

∴$U_B=\frac{mgh}{2}$

From principle of conservation of mechanical energy,

$K_0+U_0=K_B+U_B$

$\Rightarrow \frac{1}{2}m{v}_0^2+mgh=\frac{1}{2}m{v}_B^2+\frac{mgh}{2}$

We have to find $V_B$, so rearrange the equation; we can get

$v_B^2=2\times \frac{(\frac{mgh}{2}+\frac{1}{2}m{v}_0^2)}{m}$

$\Rightarrow v_B=\sqrt{2\times \frac{(\frac{mgh}{2}+\frac{1}{2}m{v}_0^2)}{m}}$

$\Rightarrow v_B=\sqrt{(gh+v_0^2)}$

$\Rightarrow v_B=\sqrt{(9.80\times 42.0+{17.0}^2)}$

$\Rightarrow v_B=\sqrt{700.6}$

$\Rightarrow v_B=26.5\text{m/s}$

The height of pointfrom bottom$=0$

∴$U_C=0$

From principle of conservation of mechanical energy,

$K_0+U_0=K_C+U_C$

$\Rightarrow \frac{1}{2}m{v}_0^2+mgh=\frac{1}{2}m{v}_C^2+0$

We have to find $v_B$, so rearrange the equation; we can get

$v_C^2=2\times \frac{(mgh+\frac{1}{2}m{v}_0^2)}{m}$

$\Rightarrow v_C=\sqrt{2\times (gh+\frac{1}{2}{v}_0^2)}$

$\Rightarrow v_C=\sqrt{(2gh+v_0^2)}$

$\Rightarrow v_C=\sqrt{(2\times 9.80\times 42.0+{17.0}^2)}$

$\Rightarrow v_C=\sqrt{1112.2}$

$\Rightarrow v_C=33.4m/s$

To find the “final” height, we set$K_f=0.$In this case, we have

$K_0+U_0=K_f+U_f$

$\Rightarrow \frac{1}{2}m{v}_0^2+mgh=\frac{1}{2}m{v}_B^2+mg{h}_f$

$\Rightarrow h_f=h+\frac{v_0^2}{2}g$

$\Rightarrow h_f=42.0+\frac{{17}^2}{2}\times 9.80$

$\Rightarrow h_f=56.7\text{m}$

From the formula, we can see that the results derived above are independent of mass. So change in mass would not change the values found above.