The labeled graphs below represent the upward trajectories of 4 bodies, two slid
ID: 1837117 • Letter: T
Question
The labeled graphs below represent the upward trajectories of 4 bodies, two sliding upwards on frictionless inclines and two in free flight. Note that all 4 bodies reach the same maximum height (9 meters) after traveling from the same initial elevation (0 m).
Answer choses are greater than, less than, or equal to
• The time of travel of P is ______ that of R
• The time of travel of N is ______ that of Q
• The initial speed of N is _______that of Q.
• The time of travel of P is ______ that of Q.
• The initial speed of R is ______ that of P.
• The initial speed of R is ______ that of Q.
Explanation / Answer
a) The time of travel of P is equal to that of R
b) The time of travel of N is less than that of Q
Time travelled by N is to reach hmax. That is motion in y direction only,
h = 0.5* g*t²
t = 2h/g
Time travelled by Q can be calculated using hchange and the y-component of acceleration;
i.e g*sin where angle of inclination to the x-axis.
t = 2h/gsin
as sin < 1 it follows that t =2h/g. So time for N is less than Q.
c)The initial speed of N is greater than that of Q.
As the intial speed of Q is utilized to give it`s K.E needed to reach 9m. But, N's initial y-component of velocity is used to give it the kinetic energy needed to reach 9m
and it also had a velocity x-component.
N has more total kinetic energy than Q. So N was faster than Q
d)The time of travel of P is less than that of Q.
As both have similar Vavg (due to uniform acceleration) but Q need still further to go.
e)The initial speed of R is greater than that of P.
f)The initial speed of R is greater than that of Q.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.