Question

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)?

Short answer

a) The speed of the car at point$A$ is$17.0\text{m/s}$

b) The speed of the car at point$B$ is$26.5\text{m/s}$

c) the speed of the car at point$C$ is$33.4\text{m/s}$

d) Final height$56.7\text{m}$

e) Even if a second car has twice the mass, it is evident that the above results do not depend on mass. Thus, a different mass for the coaster must lead to the same results.

Step-by-step solution

Given

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$v_0=17\text{m/s}$

To understand the concept

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

Formula:

$K_2+U_2=K_1+U_1$

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

$U=mgh$

(a) Calculate the speed of the car at point A

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" $\begin{array}{c}{v}_0=v_A\\ =17.0\text{m/s}\end{array}$

(b) Calculate the speed of the car at point B

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

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

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

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

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

$\Rightarrow v_B=\sqrt{700.6}$

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

(c) Calculate the speed of the car at point C

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

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

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

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

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

$\Rightarrow v_C=\sqrt{1112.2}$

$\Rightarrow v_C=33.4m/s$

(d) Calculate how high will the car go on the last hill, which is too high for it to cross

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

$K_0+U_0=K_f+U_f$

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

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

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

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

(e) Calculate the answers to (a) through and (d) if we substitute a second car with twice the mass

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.