3. (30 pts) Your little brother receives one of those rubber-ball-on-a-rubber-band toys for Christmas.
The rubber ball has a mass of 50 g and the rubber band has a spring constant of 8 N/m and is 20 cm
long when unstretched.
a) (10 pts) Holding the rubber ball initially at rest, your little brother gives it a good swat with the
wooden paddle. If the speed of the rubber ball is 5 m/s after being in contact with the wooden paddle
for 0.1 s during the swat, what average force did the paddle apply to the rubber ball?
The first way to approach this problem is through impulse. The momentum of the ball changes due to
impulse imparted by the paddle:
∆p=I=F t
∆p= (0.05)(5) −(0.05)(0) = 0.25 kg ·m/s
I= 0.25 = F(0.1)
F= 2.5 N
You could also do this by finding the average acceleration of the ball while it’s in contact with the
paddle.
a=∆v
t=5
0.1= 50 m/s2
Then use Newton’s 2nd Law to find the force required to produce this acceleration:
F=ma = (0.05)(50) = 2.5 N
b) (10 pts) Assuming the ball loses no speed until it starts to stretch the rubber band, by how much
does it stretch the rubber band before coming to rest?
∆E= ∆K+ ∆U
0 = 0−
1
2mv2+1
2kx2
−0
1
2(0.05)(5)2=1
2(8)x2
x= 0.395 m = 39.5 cm
c) (10 pts) Your brother tires of hitting the rubber ball with the wooden paddle and decides to let the
ball bounce by itself vertically at the end of the rubber band. Being an inquisitive sort, he times the
oscillations. What time should appear on the stopwatch when he measures the period of oscillation?
ω=rk
m=r8
0.05 = 12.65 rad/s
T=2π
ω=2π
12.65 = 0.523 s
