Momentum, Impulse, Collisions
Newton’s Laws
Quiz Review Practice Problems/Questions
(Reminder: please don’t forget to also enter some responses for
the Syllabus Statements related to momentum and impulse
(listed in your WebQuest worksheets). These WILL be counted in
your final score at the end of the mechanics unit when the books
are collected (mid-January). You will NOT receive a completion
stamp for them.)
Reminders
• WebQuest worksheets due tomorrow in class
• WebAssigns are due tomorrow before 1st period
• Task 3 power point (remember, rubric is online
in the WebQuest instructions!) is due tomorrow
by 11:59 PM!
• Quiz covers Newton’s laws, Momentum,
Impulse, and Conservation of Momentum
(including collisions).
Question 1
• If a 55.0 kg woman is riding in a car traveling at
90.0 km/h, what is her linear momentum
relative to:
▫ The ground?
▫ The car?
Question 1~~Answer
• If a 55.0 kg woman is riding in a car traveling at
90.0 km/h, what is her linear momentum
relative to:
▫ The ground?
 90.0km  1000m  1h 
p  mv  (55.0kg)


  1375  1380 kg m s
 1h  1km  3600s 
▫ The car?
p  mv  (55.0kg)(0.00 m s )  0 kgm s
Question 2
• A 13.5 kg lamp is hung from the ceiling using 2
cables. What is the tension in the cables if:
▫ The cables are both hung vertically?
▫ The cables are both hung so they make an angle of
63.5° with the ceiling?
Question 2~~Answer
• A 13.5 kg lamp is hung from the ceiling using 2
cables. What is the tension in the cables if:
▫ The cables are both
hung vertically?
F
y
0
FT 1  FT 2  Fg  0
2  FT  mg
(13.5kg)(9.81 m s 2 )
FT 
2
FT  66.2 N
Question 2~~Answer
• A 13.5 kg lamp is hung from the ceiling using 2
cables. What is the tension in the cables if:
▫ The cables are both hung so they make an angle of
63.5° with the ceiling?
F
y
0
2  FTy  mg
2  FT  Sin (63.5 )  (13.5kg)(9.81 m s 2 )
(132.4 N )
FT 
 74.0 N

2  Sin (63.5 )
Question 3
• How fast would a 1160 kg car travel (in m/s)if it
had the same linear momentum as a 1570 kg
truck traveling at 90.0 km/h?
Question 3~~answer
• How fast would a 1160 kg car travel (in m/s)if it
had the same linear momentum as a 1570 kg
truck traveling at 90 km/h?
vtruck  (90.0 km h )(1000 m km)(1h
)
3600s
vtruck  25.0 m s
pcar  ptruck
(m  v) car  (m  v) truck
(1160kg)(v)  (1570kg)( 25.0 m s )
vcar
39250

 33.8 m s
1160
Question 4
• A 0.150 kg baseball traveling with a horizontal
speed of 5.00 m/s is hit by a bat and then moves
with a speed of 42.4 m/s in the opposite
direction. What is the change in the ball's
momentum?
Question 4~~Answer
• A 0.150 kg baseball traveling with a horizontal
speed of 5.00 m/s is hit by a bat and then moves
with a speed of 42.4 m/s in the opposite
direction. What is the change in the ball's
momentum? p  m  v
p  m  (v f  vi )
p  (0.150kg)( 42.4 m s  (5.00 m s ))
p  (0.150)  (47.4)
p  7.11 kg m s
Question 5
• A 15.0 g rubber bullet hits a wall with a speed of
150 m/s. If the bullet bounces straight back with
a speed of 125 m/s, what is the change in
momentum of the bullet?
Question 5~~Answer
• A 15.0 g rubber bullet hits a wall with a speed of
150 m/s. If the bullet bounces straight back with
a speed of 125 m/s, what is the change in
momentum of the bullet?
p  m  v
p  m  (v f  vi )
p  (0.015kg)(125 m s  (150 m s ))
p  (0.015)  (275)
p  4.13 kg m s
Question 6
• A person pushes a 12.3 kg box from rest and
accelerates it to a speed of 2.17 m/s with a
constant force. If the box is pushed for a time of
2.35 s, what is the force exerted by the person?
Question 6~~Answer
• A person pushes a 12.3 kg box from rest and
accelerates it to a speed of 2.17 m/s with a
constant force. If the box is pushed for a time of
2.35 s, what is the force exerted by the person?
F  t  m  v
F (2.35s)  (12.3kg)( 2.17 m s )
F  11.4 N
Question 7
• A loaded tractor-trailer with a total mass of
3200 kg traveling at 2.0 km/h hits a loading
dock and comes to a stop in 0.64 s. What is the
magnitude of the average force exerted on the
truck by the dock?
Question 7~~Answer
• A loaded tractor-trailer with a total mass of
3200 kg traveling at 2.0 km/h hits a loading
dock and comes to a stop in 0.64 s. What is the
magnitude of the average force exerted on the
truck by the dock?
F  t  m   v
F (0.64 s )  (3200kg)( 2.0 km h )(1000 m km )(1h 3600s )
F (0.64)  1777.8
F  2777.8 N  2800 N
Question 8
• A 1.32 kg mud ball drops from rest at a height of
10.0 m. If the impact between the ball and the
ground lasts 0.604 s, what is the average net
force exerted by the ball on the ground?
Question 8 ~~Answer
• A 1.32 kg mud ball drops from rest at a height of
10.0 m. If the impact between the ball and the
ground lasts 0.604 s, what is the average net
force exerted by the ball on the ground?
d  10.0m
v 2  v0  2ad
a  9.81 m s 2
v 2  0 2  2(9.81)(10.0)
v0  0
v  196.2  14.0 m s
2
v?
F  t  m  v
m  1.32kg
F (0.604s)  (1.32kg)(14.0 m s )
t  0.604 s
F ?
F  30.6 N
Question 9
• A 0.5 kg ball is thrown horizontally with a
velocity of 15 m/s against a wall. If the ball
rebounds horizontally with a velocity of 13 m/s
and the contact time is 0.025 s, what is the force
exerted on the ball by the wall?
Question 9~~Answer
• A 0.5 kg ball is thrown horizontally with a
velocity of 15 m/s against a wall. If the ball
rebounds horizontally with a velocity of 13 m/s
and the contact time is 0.025 s, what is the force
exerted on the ball by the wall?
𝐹 ∙ ∆𝑡 = 𝑚 ∙ ∆𝑣
𝐹 ∙ 0.025𝑠 = 0.50𝑘𝑔 ∙ 13 𝑚 𝑠 − −15 𝑚 𝑠
𝑘𝑔𝑚
14
𝑠
𝐹=
= 560𝑁
0.025𝑠
Question 10
• A one-dimensional impulse force acts on a 4.50 kg
object as diagrammed in the figure below.
• (a) What is the magnitude
of the impulse given to the
object?
(b) What is the magnitude
of the average force?
(c) What is the final velocity if the object had an
initial speed of 5.80 m/s in a direction opposite to
the impulse force?
Question 10~~Answer
• A one-dimensional impulse force acts on a 4.50
kg object as diagrammed in the figure below.
• (a) What is the magnitude
of the impulse given to the
object?
Impulse = area under the line
Impulse = (12(𝑏1+𝑏2 )) ∙ ℎ
Impulse = (0.5)(0.14+(0.08-0.05))·900
• Impulse = 76.5 N·s
Question 10~~Answer
• A one-dimensional impulse force acts on a 4.50
kg object as diagrammed in the figure below.
• (b) What is the magnitude
of the average force?
Impulse = F·t
(76.5 N·s) = F·(0.14s)
546 N = F
F = 550 N
Question 10~~Answer
• A one-dimensional impulse force acts on a 4.50
kg object as diagrammed in the figure below.
(c) What is the final velocity if the object had an
initial speed of 5.80 m/s in a direction opposite
to the impulse force?
F·t=m·v
(76.5 N·s)=(4.50 kg)(vf – (-5.80m/s))
76.5
= 𝑣𝑓 + 5.80
4.50
17.0 − 5.80 = 𝑣𝑓 = 11.2𝑚/𝑠
Question 11
• A 16000 N automobile travels at a speed of 50 km/h
northward along a street, and a 9000 N sports car
travels at a speed of 64 km/h eastward along an
intersecting street.
(a) If neither driver brakes and the cars collide at the
intersection and lock bumpers, what will the velocity
of the cars be immediately after the collision?
(magnitude AND direction)
Question 11~~Answer
• A 16000 N automobile travels at a speed of 50 km/h
northward along a street, and a 9000 N sports car
travels at a speed of 64 km/h eastward along an
intersecting street.
(a) If neither driver brakes and the cars collide at the
intersection and lock bumpers, what will the velocity
of the cars be immediately after the collision?
Step 1: determine the mass of each car
Step 2: determine the momentum of each car
step 3: draw your vector diagram! Each car is a
component of the total final momentum
Question 11~~Answer
• A 16000 N automobile travels at a speed of 50 km/h
northward along a street, and a 9000 N sports car
travels at a speed of 64 km/h eastward along an
intersecting street.
(a) If neither driver brakes and the cars collide at the
intersection and lock bumpers, what will the velocity
of the cars be immediately after the collision?
m1=16000 N/9.81 m·s-2=1631 kg
m2=9000 N/9.81 m·s-2=917 kg
p1=m·v=(1631 kg)·(50km/h)(1000m/km)(1h/3600s)
p1=22650 Ns
p2=m·v=(917 kg)·(64km/h)(1000m/km)(1h/3600s)
p2=16302 Ns
Question 11
• A 16000 N automobile travels at a speed of 50 km/h
northward along a street, and a 9000 N sports car
travels at a speed of 64 km/h eastward along an
intersecting street.
(a) If neither driver brakes and the cars collide at the
intersection and lock bumpers, what will the velocity
of the cars be immediately after the collision?
16302 Ns
22650 Ns
p(final)=
226502 + 163022 = 27900 𝑁𝑠
16302
)=
22650
𝜃 = 𝑡𝑎𝑛−1 (