South Pasadena • A.P. Physics
Name
7 · Momentum
Period
7.1
WORKSHEET
–
Date
MOMENTUM AND IMPULSE
1. An 8-kg bowling ball is rolling at 2 m/s.
(a)
What is the momentum of the bowling ball?
m = 8 kg
v = 2 m/s
p = mv = (8 kg)(2 m/s) = 16 N·s
(b)
If the bowling ball rolls into a pillow and stops in 0.5 s, calculate the average force it
exerts on the pillow.
m = 8 kg
m(v – v0) (8 kg)(0 m/s – 2 m/s)
v0 = 2 m/s
Ft = m(v – v0)  F =
=
= –32 N
t
0.5 s
v = 0 m/s
t = 0.5 s
What average force does the pillow exert on the ball?
Fbp = – Fpb = 32 N
(c)
2. Assume an 8-kg bowling ball moving at 2 m/s bounces off a spring at the same speed that it had before
bouncing.
(a) What is its momentum of recoil?
m = 8 kg
v = –2 m/s
p = mv = (8 kg)(–2 m/s) = –16 N·s
(b)
What is its change in momentum?
m = 8 kg
v0 = 2 m/s
∆p = p – p0 = mv – mv0 = m(v – v0) = (8 kg)(–2 m/s – 2 m/s)
v = –2 m/s
= –32 N·s
(c)
If the interaction with the spring occurs in 0.5 s, calculate the average force the spring
exerts on it.
∆p = –32 N·s
∆p –32 N·s
t = 0.5 s
F=
=
= –64 N
t
0.5 s
(d)
Compare this force with the force from question #1.
This force is twice as large as the force in #1.
3. A 50-kg carton that slides at 4 m/s across an icy surface
(a)
What is the momentum of the carton?
m = 50 kg
v = 4 m/s
p = mv = (50 kg)(4 m/s) = 200 N·s
(b)
The sliding carton skids into a rough surface and stops in 3 s. Calculate the
force of friction it encounters.
m = 50 kg
m(v – v0) (50 kg)(0 m/s – 4 m/s)
v0 = 4 m/s
F=
=
= –67 N
t
3s
v = 0 m/s
t=3s
4. A force of 10 N is exerted on a cart for 2.5 s.
(a)
What is the impulse of the cart?
F = 10 N
t = 2.5 s
J = Ft = (10 N)(2.5 s) = 25 N·s
(b)
What change in momentum does the cart undergo?
J = 25 N·s
∆p = J = 25 N·s
(c)
If the mass of the cart is 2 kg and the cart is initially at rest, calculate its final
speed.
∆p = 25 N·s
∆p = m(v – v0)
∆p
25 N·s
m = 2 kg
v=
+ v0 =
+ 0 m/s = 12.5 m/s
m
2 kg
v0 = 0 m/s
5. A car with a mass of 1000 kg moves at 20 m/s. What braking force is needed to
bring it to a halt in 6 s?
m = 1000 kg
m(v – v0) (1000 kg)(0 m/s – 20 m/s)
v0 = 20 m/s
F=
=
= 3333 N
t
6s
v = 0 m/s
t=6s