On a snow day from Fayetteville Public Schools many years ago, Kat and her frien
ID: 1588045 • Letter: O
Question
On a snow day from Fayetteville Public Schools many years ago, Kat and her friend Armon (40 kg) decided to go sledding on their street, which had been closed. Assume the sled is very slick, and Armon barely needed to nudge off from rest when he began to slide down the 5 hill. Kat (25 kg) was skipping down the hill at 1 m/s. Armon caught up with her 5.0 m from where he started, scooped her smoothly onto the sled and continued down the hill. (a)[4 pts] What was Armon’s acceleration as he slid down the hill? (b)[6 pts] How much time did it take for him to catch Kat? (c)[5 pts] How fast was he going the instant before he caught Kat? (d)[4 pts] What was Armon’s momentum the instant before he caught Kat? (e)[5 pts] In order for total momentum to be constant for the collision of Armon and Kat, what assumptions must you make? Why? (f)[8 pts] What was Armon and Kat’s new velocity immediately after that collision?
Explanation / Answer
Ans:a)Arman acceleration down the hill is a =g*sin(theta); here g=9.8 m/sec^2 and theta =5 degree; so a=0.85 m/sec^2.
b)Arman velocity at time t is v = a*t as he starts from rest so initial velocity was zero. And distance travel in time t by Arman is d= 0.5*a*t^2 ; Now d=5 m so t=sqrt(2d/g) = 3.43 sec.
c)So Arman velocity at the time of caught v= 2.92 m/sec.
d) Armam momentum just be he caught Kat is pA= m*v= 40*2.92 =116.33 kg-m/sec
e)In order for total momentum to be constant for the collision of Armon and Kat,we assumed that they collide elastically and their momrntum adds up after collision.
f) Noe Kat momentum = 25*1=25 kg-m/sec
if P=total momentum after collision= Arman momentum + Kat momentum=116.3 +25 =141.3 kg-m/sec
velocity after collision V= P/(total mass) =141.3/65 =2.17 m/sec