91影视

i) Mass of car$m=825\text{鈥嬧赌塳驳}$

ii) The height $h=42\text{鈥尘}$

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,

鈭�$U_A=U_0$

$K_0+U_0=K_A+U_A$

鈭�$K_0=K_A$

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

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

鈭�$U_B=\frac{mgh}{2}$

From principle of conservation of mechanical energy,

$K_0+U_0=K_B+U_B$

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

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

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

$鈬�v_B=\sqrt{2脳\frac{(\frac{mgh}{2}+\frac{1}{2}m{v}_0^2)}{m}}$

$鈬�v_B=\sqrt{(gh+v_0^2)}$

$鈬�v_B=\sqrt{(9.80脳42.0+{17.0}^2)}$

$鈬�v_B=\sqrt{700.6}$

$鈬�v_B=26.5\text{m/s}$

The height of pointfrom bottom$=0$

鈭�$U_C=0$

From principle of conservation of mechanical energy,

$K_0+U_0=K_C+U_C$

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

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

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

$鈬�v_C=\sqrt{2脳(gh+\frac{1}{2}{v}_0^2)}$

$鈬�v_C=\sqrt{(2gh+v_0^2)}$

$鈬�v_C=\sqrt{(2脳9.80脳42.0+{17.0}^2)}$

$鈬�v_C=\sqrt{1112.2}$

$鈬�v_C=33.4m/s$

To find the 鈥渇inal鈥� height, we set$K_f=0.$In this case, we have

$K_0+U_0=K_f+U_f$

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

$鈬�h_f=h+\frac{v_0^2}{2}g$

$鈬�h_f=42.0+\frac{{17}^2}{2}脳9.80$

$鈬�h_f=56.7\text{m}$

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.