Finding distance from velocity, force, momentum, mass, and time

[ari-anne]

Homework Statement

Two skaters, initially at rest, push on each other for 1.5 s with a force of 60 N. Ignore the effects of friction. The skaters have mass 45 kg and 70 kg.
(a) What is the speed of the 45 kg skater after the push?
(b) What is the speed of the 70 kg skater after the push?
(c) Over what distance did the 60 N force act on the 70kg skater?

Relevant Equations

impulse=Ft
impulse=change in p
p=mv
F=change in p/t
possiblt more?

I solved a and b fine, I just don't know where to start c
a) i=Ft i=change in p and p=mv so i=90=45v so v=2m/s
b) same thing just using the other mass i=90=70v so v=1.3m/s
c) v=d/t so 1.3=d/1.5 but that would be d=1.95 and the answer key says d=.96m

 

[gneill]

Hi ari-anne,

Welcome to Physics Forums!

For part (c) notice that the skater will being accelerated over the time that the force is acting. So a simple v = d/t won't cut it here.

Can you think of a SUVAT equation that relates velocity to acceleration and distance?

 

[ari-anne]

Oh so would I use d=vt-1/2at^2 with a =.86m/s^2
d=1.3 x 1.5 - .5 x .86 x 1.5^2
d=1.95-.97= .98 and just chalk the .02 difference to rounding or am I still doing something wrong?

 

[gneill]

You should definitely not round intermediate results. Keep more digits in intermediate results, only rounding at the very end.

In this case your initial velocity starts out at zero, so the final velocity reached is

$v_f = \frac{1}{2} a t^2$

I might have gone with the SUVAT equation

$v_f^2 = 2 a d$

solving for d.

 

[Orodruin]

I would start from $Fd = p^2/(2m)$, but maybe that is just me ... The result should of course be the same.

 

[DEvens]

The way to deal with rounding in this sort of question is to leave the equations in symbols until the very end. Only when you report numerical values should you put the numbers in. And when you report an intermediate value, such as part b, don't use that rounded value in later calculations. Instead, use the symbols again.

Also, in this sort of question, you could consider leaving the numbers as fractions. So instead of writing 1.3 where you did, you could have written 90/70=9/7.