Question

A 77.0 kg rider sitting on a 7.3 kg bike is riding along at 9.3 m/s in the positive direction. The rider drags a foot on the ground and slows down to 6.0 m/s still in the positive direction.


A) What is the change in momentum of the rider and bike?

B) What is the impulse delivered by the ground to the rider's foot?

C) What force is acting on the bike and rider if slowing down took 13.2 seconds?

D) And, how far did the bike and rider travel during these 13.2 seconds?

Answer 1

Answer with Explanation:

We are given that

Mass of rider=m=77 kg

Mass of bike =m'=7.3 kg

Initial velocity,u=9.3 m/s

Final velocity,v=6 m/s

A.Change in velocity=v-u=6-9.3=-3.3 m/s

Total mass,M=m+m'=77+7.3=84.3 kg

Change in momentum=$M(v-u)=84.3(-3.3)=-278.19 kgm/s$

B.Impulse=Ft=Change in momentum=-278.19kg m/s

C.Time,t=13.2 s

$v=u+at$

Using the formula

$6=9.3+13.2a$

$13.2a=6-9.3=-3.3$

$a=-\frac{3.3}{13.2}=-0.25 ms^{-2}$

F=$ma$

$F=84.3(-0.25)=-21.075 N$

D.$S=ut+\frac{1}{2}at^2$

$S=9.3(13.2)+\frac{1}{2}(-0.25)(13.2)^2$

$S=100.98 m$