Answer to Question #286864 in Mechanics | Relativity for Collins

Question

Dancers have to learn many skills including how land correctly.A dancer of mass 50 kg leaps into the air and lands feet first on the ground .She lands on the ground with a velocity of 5metres per second.As she lands,she bends her knees and comes to a complete stop in 0,2 seconds.

A. calculate the momentum with which the dancer reaches the ground

B. Define impulse

C. Calculate the magnitude of the net force acting on the dancer as she lands​

Answer 1

Explanations & Calculations

• Assume the dancer lands vertically on the ground so that the velocity given can be treated without performing any resolution.

a)

• Momentum at any given point is $\small mv.$ Therefore, the momentum at the landing is $\small 50\,kg\times5\,ms^{-1}=250\,kgms^{-1}$ .
• Always remember to use units as much as possible.

b)

• Impulse is the change of momentum within a certain period of time.
• It has the same units as the momentum.

c)

• That impulse generated during 0.2 seconds delivers a force on her.
• That force can be calculated considering the impulse per unit time.

$\qquad\qquad \begin{aligned} \small |F|&=\small\frac{I}{\Delta t}=\frac{\Delta mv}{\Delta t}\\ &=\small \frac{50\,kg|(0-5\,ms^{-1})|}{0.2\,s}\\ &=\small 1250\,N \end{aligned}$

• This is only due to the change of momentum.
• The net force is this plus the body weight which is,

$\qquad\qquad \begin{aligned} \small F_{net}&=\small 1250+50g\\ &=\small1250+490\\ &=\small1740\,N \end{aligned}$