Space Time & Motion A.2 Forces & Momentum (id: 0b94cd5e9)

本题目来源于试卷: Space Time & Motion A.2 Forces & Momentum，类别为 IB物理学

[单选题]

A cart of mass mm enters a smooth vertical loop the loop or radius R with a horizontal speed u at the bottom of the track, as shown. The cart just manages to stay in contact with the track at the top most point of its path.

Its speed at the bottom of the loop u, is given by

:

A. $u=\sqrt5gR$
B. $u=\sqrt4gR$
C. $u=\sqrt1gR$
D. $u=\sqrt2gR$


参考答案:  A


本题详细解析:

$\underline{\text{Explanation}}$: 
Conservation of energy: 
$E_k+E_p$ at the bottom of the circle = $E_k+E_p$ at the top of the circle.

(I)$$ $\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}^2+mg(2R)⟹u^2=v^2+4gR$

Centripetal force at the top of the loop comes from the weight of the cart and the contact force from the track. Writing Newton’s second law at that point gives:

$\begin{array}{rl}{\displaystyle mg+N}& {\displaystyle =m(\frac{v^2}{R})}\\ {\displaystyle }\\ {\displaystyle mg+0}& {\displaystyle =m(\frac{m{v}^2}{R})}\\ {\displaystyle }\\ {\displaystyle v^2}& {\displaystyle =gR}\end{array}$

Substituting this result into (I) gives $u^2=gR+4gR$, or $u=\sqrt{5gR}$

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