Question

An object moving at a constant speed of 29 m/s is making a turn with a radius of curvature of 9 m (this is the radius of the "kissing circle"). The object's momentum has a magnitude of 61 kg·m/s. What is the magnitude of the rate of change of the momentum?

Answer 1

Answer:

The rate of change of the momentum is 196.5 kg-m/s²

Explanation:

Given that,

Speed v = 29 m/s

Radius = 9 m

Momentum = 61 kg-m/s

We need to calculate the rate of change of the momentum

Using formula of momentum

$F = \dfrac{\Delta p}{\Delta t}$

$F=\dfrac{\Delta(mv)}{\Delta t}$.....(I)

Using newtons second law

$F = ma=m\dfrac{v^2}{r}$

$F = \dfrac{mv(v)}{r}$....(II)

From equation (I) and (II)

$\dfrac{\Delta p}{\Delta t}=\dfrac{61\times29}{9}$

$\dfrac{\Delta p}{\Delta t}=196.5\ kg-m/s^2$

Hence, The rate of change of the momentum is 196.5 kg-m/s²