Consider two identical balls of mass M and radius R rolling down inclines that a
ID: 1494960 • Letter: C
Question
Consider two identical balls of mass M and radius R rolling down inclines that are identical in each way except that one of them is frictionless and one is not. Ball A slides down its ramp, and Ball B rolls without slipping. At the bottom of the ramp both balls strike identical frictionless, horizontal springs, and compress them some distance before coming to rest. If the balls are hollow spheres with mass M=0.5 kg and radius R=0.25 m, the ramp is 3 meters tall, and the springs both have a spring constant k=5 N/m, how far does each ball compress their respective spring? Be sure to specify what condition(s) allow you to use the method that you have chosen.Explanation / Answer
mass = M = 0.5 kg
radius = R = 0.25 m
for frictionless ball
PE at top of incline = mgh = 0.5*9.8*3 = 14.7 J
KE at bottom of incline = PE at top of incline [ conse3rvation of enerrgy]
Nowm, this KE is used to compress the spring
14.7 = 0.5kx^2 [ where x is the compression of the spring]
14.7*2/5 = x^2
x = 2.42487 m
for friction ball
PE at top of incline = mgh = 14.7 J
Rotational KE at bottom of incline = 0.5*I*w^2
for hollow sphere , I = 2mr^2/3
w = v/r [for rolling]
so rotational KE = 0.5*2*mv^2/3 = mv^2/3
Translational KE = 0.5mv^2
Net KE = 0.8333mv^2 = 14.7
v^2 = 14.7/0.8333*0.5
v = 5.93 m/s
translational ke = 0.5mv^2 = 8.82 J
translational ke is used to compress the spring, 8.82 = 0.5*5*x^2
x = 1.8782 m