Question

A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and the speed is 10 m/s. At this instant the cord is 3.40 m long. (Take the x direction to be horizontal and to the right, the y direction to be upward, and the z direction to be out of the page.)
At this instant, what is the magnitude of the rate of change of the child's momentum?

Answer 1

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.