# Impulse integration for a Tennis Racket hitting a Tennis Ball

• ChiralSuperfields
In summary, the conversation discusses the integration of impulse over time and the use of the endpoints or the function itself as the bounds. It is noted that if the impulse is a function of time, then integrating from ##t_i## to ##t_f## is reasonable. However, omitting the "t=" from the bounds can be misleading.
ChiralSuperfields
Homework Statement
Relevant Equations
For this,

Can someone please tell me why they integrate the impulse over from ##t_i## to ##t_f##? Why not from ##j_i## to ##j_f##? It seems strange integrating impulse with respect to time.

Many thanks!

ChiralSuperfields said:

For this,
View attachment 326569
Can someone please tell me why they integrate the impulse over from ##t_i## to ##t_f##? Why not from ##j_i## to ##j_f##? It seems strange integrating impulse with respect to time.

Many thanks!
If you are using
##\displaystyle \int d \textbf{J}##
and you have the ##\textbf{J}##'s at the endpoints, the use the ##\textbf{J}##'s.

If you don't have the ##\textbf{J}##'s then you need to use
##\displaystyle \int \textbf{F}(t) \, dt##

-Dan

ChiralSuperfields
ChiralSuperfields said:
Can someone please tell me why they integrate the impulse over from ##t_i## to ##t_f##? Why not from ##j_i## to ##j_f##? It seems strange integrating impulse with respect to time.
If you think of ##\vec J## as the cumulative impulse given over some period of time then ##\vec J=\vec J(t)## and it is reasonable to write ##\int_{t=t_i}^{t_f}d\vec J(t)##. But omitting the "t=" from the bounds is a bit naughty.

topsquark and ChiralSuperfields
haruspex said:
If you think of ##\vec J## as the cumulative impulse given over some period of time then ##\vec J=\vec J(t)## and it is reasonable to write ##\int_{t=t_i}^{t_f}d\vec J(t)##. But omitting the "t=" from the bounds is a bit naughty.

@haruspex, now that you say J is a function of t I think that helps.

Many thanks!

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