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Physics-

General

Easy

Question

Consider a particle moving without friction on a rippled surface, as shown. Gravity acts down in the negative h direction. The elevation h(x) of the surface is given by h(x) = d cos(kx). If the particle starts at x = 0 with a speed v in the x direction, for what values of v will the particle stay on the surface at all times?

$v \leq \sqrt{g d}$
$v \leq \sqrt{\frac{g}{k}}$
$v \leq \sqrt{g k d^{2}}$
$v \leq \sqrt{\frac{g}{k^{2} d}}$

The correct answer is: $v less or equal than square root of fraction numerator g over denominator k to the power of 2 end exponent d end fraction end root$

$table row cell R subscript C end subscript equals fraction numerator open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 3 divided by 2 end exponent over denominator d to the power of 2 end exponent y divided by x to the power of 2 end exponent end fraction comma blank h e r e blank R subscript C end subscript blank a t blank x equals 0 equals fraction numerator 1 over denominator k to the power of 2 end exponent d end fraction end cell row cell therefore v less or equal than square root of g R subscript C end subscript end root equals square root of fraction numerator g over denominator k to the power of 2 end exponent d end fraction end root end cell end table$

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