Name: _________________________________ Date: ________ Period: ________
Mr. Talboo – Physics
Momentum and Impulse SAMPLE PROBLEMS
SAMPLE A
A 30000-N truck is traveling 50 km/h West.
(a) Find the momentum for this truck.
p = mv = (3061.224)(13.889)
= 42517.340 kg · m/s
(b) Calculate the impulse imparted on the truck if it brakes to a stop in 35 s.
I = m∆v = (3061.224)(-13.889) = -42517.340 N·s
(c) Calculate the force necessary to stop the truck.
F = I / ∆t = (-42517.340) / (35) = -1214.781 N
SAMPLE B
Bam Margera accelerates his Lamborghini Murcielago
Roadster from 70 m/s to 150 m/s in a time of 2.8 s.
(a) What is the car’s change in momentum if it’s mass
is 1500 kg?
∆p = mvf – mvi = = (1500)(150) - (1500)(70)
= 120000 kg · m/s
(b) What impulse was imparted on the car to cause this acceleration?
I = ∆p = 120000 N · s
(c) If the time required to accelerate the car shortened, would the impulse
increase or decrease?
The same
SAMPLE C
In 2001, Dale Earnhardt’s 1545-kg car crashed into
the wall at Daytona International Speedway. In
doing so, Dale experienced a velocity change of 19
m/s for a time of approximately 80 milliseconds.
(a) What was the car’s momentum change?
∆p = m∆v = = (1545)(-19) = -29,355 kg · m/s
(b) What was the impulse delivered to the car?
I = ∆p = -29,355 N · s
(c) Calculate the force of impact responsible for this momentum change.
F = I / ∆t = -29,355 / 0.08 = -366,937.5 N
SAMPLE D
In a particular crash test, an
automobile of mass 1500-kg collides
with a wall. The initial and final
velocities of the automobile are vi = 15.0 m/s and vf = 2.60 m/s. If the
collision lasts for 0.150 s, find the
impulse imparted on the automobile
and the force of impact.
I = m∆v = = (1500)(-17.6) = 26,400 N · s
F = I / ∆t = 26,400 / 0.15 = 176,000 N
SAMPLE E
A 7 kg bowling ball rolls into a 2.27 kg pin in
an elastic collision. The bowling ball hit the pin
with an initial velocity of 8.3 m/s. If the final
velocity of the bowling ball is 7.2 m/s, what is
the final velocity of the pin?
m1v1i + m2v2i = m1v1f + m2v2f
(7 · 8.3) + (2.27 · 0) = (7 · 7.2) + (2.27 · vf)
58.1
+
0
-50.4
7.7 = 2.27(vf)
2.27 2.27
vf = 3.392 m/s
= 50.4
-50.4
+ 2.27(vf)
SAMPLE F
A 730 kg Smart For Two is sitting at a stoplight. A 5919 kg Ford F450
makes a totally inelastic collision with the Smart Car at a speed of 20 m/s.
a) What is the final velocity of the Smart Car and the F450?
m1v1i + m2v2i = vf
(m1+ m2)
vf
=
(5919 · 20) + (730 · 0)
(5919 + 730)
=
17.804 m/s
b) If the Smart Car hit a stationary F450 at the same speed, what would the
final velocity be?
m1v1i + m2v2i = vf
(m1+ m2)
vf = (5919 · 0) + (730 · 20) = 2.196 m/s
(5919 + 730)
c) Why are these two velocities so different?
The Ford has much more momentum than the
Smart Car.