Question

A cart rolls halfway down a flat, frictionless ramp so that none of the mechanical energy (kinetic plus potential) is lost. How much speed does the cart have halfway down($v_1$), compared to how much it will have at the bottom ($v_2$)?

• A. $v_1=0.5v_2$
• B. $v_1<0.5v_2$
• C. $0.5v_2<v_1<0.75v_2$
• D. $v_1=0.25v_2$
• E. $v_1=0.75v_2$

Answer 1

The correct option is C $0.5v_2<v_1<0.75v_2$

The total mechanical energy is conserved,

Let the speed halfway down be $v_1$ and that at the bottom be $v_2$.

Speed halfway down is given by

$\frac{1}{2}m{v}_1^2=mgh$

$⟹v_1=\sqrt{2gh}$

The speed at the bottom is given by-

$\frac{1}{2}m{v}_2^2=mg\left(2h\right)$

$⟹v_2=\sqrt{4gh}$

$⟹v_2=\sqrt{2}{v}_1$

$⟹v_1=\frac{v_2}{\sqrt{2}}\approx 0.707v_2$