Chapter 1 6 Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #) 978-049511374-3
THE PROBLEM STATEMENT
Ch 1 6. Kinetic energy KE ( = ½ mv2 , = ½ mass . velocity2 ; see Chapter 5) has dimensions
kg· m2/s2 (mass . meter2 /second2). It can be written in terms of the momentum p (Chapter 6) and
mass m as
KE= p2 / 2m .
(a) Determine the proper units for momentum using dimensional analysis.
(b) Refer to Problem 5. Given the units of force, write a simple equation relating a constant force
F exerted on an object, an interval of time t during which the force is applied, and the
resulting momentum of the object, p.
Page 1 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 1 6. Kinetic energy KE ( = ½ mv2 , = ½ mass . velocity2 ; see Chapter 5) has dimensions
kg· m2/s2 (mass . meter2 /second2). It can be written in terms of the momentum p (Chapter 6) and
mass m as
KE= p2 / 2m .
(a) Determine the proper units for momentum using dimensional analysis.
(b) Refer to Problem 5. Given the units of force, write a simple equation relating a constant force
F exerted on an object, an interval of time t during which the force is applied, and the
resulting momentum of the object, p.
BRAINSTORMING - Definitions, concepts , principles and Discussion
KE = ½ mv2
Units
m
t
x
v
a
F = ma ,
Physical Dimensions
M
s
T
m
L
m/s
L/T
L/T2
m/s2
kg*m/s2 ML/T2
KE
kg m2/s2
kg
ML2/T2
.
p = mv
p
kg m/s
ML/T
ONLY DONE AS AN ILLUSTRATIVE EXAMPLE Let’s play with KE = kg m2/s2 = ML2/T2.
Notice it is (M L/T)*(L/T) which is p*(L/T) because p is ML/T.
But, if L/T is multiplied and divided by M , we get (M L/T)/M. This is p/M .
So now KE = M/L2/T2 = (ML/T)*(ML/T)/ M = p*p/ kg = p2/kg = p2/m Hence KE has the units as p2/m
Therefore KE = ½ p2/m .
USE THIS METHOD FOR a. and b.
Page 2 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 1 6. Kinetic energy KE ( = ½ mv2 , = ½ mass . velocity2 ; see Chapter 5) has dimensions
kg· m2/s2 (mass . meter2 /second2). It can be written in terms of the momentum p (Chapter 6) and
mass m as
KE= p2 / 2m .
(a) Determine the proper units for momentum using dimensional analysis.
(b) Refer to Problem 5. Given the units of force, write a simple equation relating a constant force
F exerted on an object, an interval of time t during which the force is applied, and the
resulting momentum of the object, p.
Basic Solution
Units
a.
b.
m
t
x
v
a
F = ma ,
Physical Dimensions
kg
M
s
T
m
L
m/s
L/T
2
L/T2
m/s
kg*m/s2 ML/T2
KE
kg m2/s2
p = mv
= mv = kg m/s = M L/T
Units
Dimensions
Show that p = F t
We know that p = (ML/T) and
Since
ML2/T2
F = (ML/T2).
p = ML/T , we need a T2 on the bottom to get F.
So multiplying and dividing p by T, gives pT/T = (ML/T) *T/T .
Putting the T on the bottom (denominator) into the brackets, ( ) , yields
p = (ML/(T*T))T = (ML/T2) T.
But F = (ML/T2), so we have
p =
(ML/T2) T = FT .
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