, 1
2.X.1 Let ~pi = h1.5, −1.2, 0.7i kg · m/s and ~pf = h1.6, −0.9, 1.1i kg · m/s. The change in momentum is just ~pf - ~pi .
∆~p = ~pf − ~pi
= h1.6, −0.9, 1.1i kg · m/s − h1.5, −1.2, 0.7i kg · m/s
= h0.1, 0.3, 0.4i kg · m/s
2.X.2 The net force on a system is always the vector sum, or superposition, of the individual forces acting on the system.
~F = h40, −70, 0i N + h20, 10, 0i N
net,sys
= h60, −60, 0i N
2.X.3 This problem is best solved using simple ratio reasoning. Changing the force’s magnitude by a factor of two, for the
same duration, must also change the resulting change in momentum. Halving the force must halve the change in momentum.
So |~p| = 1.5 kg · m/s.
2.X.4 Let the object (whatever it is) be the system. We’re given that ~Fnet,sys = h−0.5, −0.2, 0.8i N and ∆t = 2 min = 120 s
(don’t forget to express ∆t in seconds!).
(a) We have to calculate the impulse as follows:
∆~p = ~Fnet,sys ∆t
= (h−0.5, −0.2, 0.8i N) (120 s)
= h−60, −24, 96i kg · m/s
Note that this is (must be!) the same value we would have gotten had we known both ~pf and ~pi .
(b) Remember that impulse and change in momentum are merely two different words for the same quantity. This quantity
can be called impulse if it is calculated from the net force and its duration. It can also be called change in momentum
if it is calculated from the initial and final momenta. Names aside, it is the same quantity in both cases!
2.X.5 Let the system consist of the hockey puck. Assume there are no significant interactions on the system other than from
ice and air (we can treat them as one net interaction), and this interaction was constant while it existed. We’re given that
~pi = h0, 2, 0i kg · m/s.
(a) If the puck comes to a stop, its final momentum must be zero. Therefore ~pf = h0, 0, 0i kg · m/s. The impulse, or change
in momentum, is merely ∆~p = ~pf - ~pi = h0, 0, 0i kg · m/s - h0, 2, 0i kg · m/s = h0, −2, 0i kg · m/s.
(b) We know ∆~p and ∆t so we can calculate ~Fnet,sys using the momentum principle.
∆~p = ~Fnet,sys ∆t
~F ∆~p
=
net,sys
∆t
~F h0, −2, 0i kg · m/s
=
net,sys
3
s
2
~F = 0, − , 0 N
net,sys
3
,2
2.X.6 This is a straightforward application of the momentum principle. Let the system be the puck, and assume the only
significant interaction on the system is with the hockey stick.
∆~p = ~Fnet,sys ∆t
~pf − ~pi = ~Fnet,sys ∆t
~pf = ~pi + ~Fnet,sys ∆t
~pf = h10, 0, 5i kg · m/s + h0, 0, 2000i N × (0.004 s)
~pf ≈ h10, 0, 13i kg · m/s
Note that the puck’s x and y components of momentum didn’t change because the net force had no components in those
directions.
2.X.7 We’re given that ~vi = h25, 0, 15i m/s, ~vf = h10, 0, 18i m/s, and mcar = 1000 kg. Let the system consist of the car,
and let’s assume non-relativistic speeds.
(a)
∆~psys = ~pf − ~pi
∆~psys ≈ mcar ~vf − mcar ~vi
∆~psys ≈ mcar ~vf − mcar ~vi
∆~psys ≈ (1000 kg) (h10, 0, 18i m/s − h25, 0, 15i m/s)
∆~psys ≈ (1000 kg) (h−15, 0, 3i m/s)
D 4 3
E
∆~psys ≈ −1.5 × 10 , 0, 3 × 10 kg · m/s
(b) Remember that impulse and change in momentum refer to the same physical quantity, so the answer is the same as
above.
(c)
∆~psys
~F =
D∆t
net,sys
4 3
E
−1.5 × 10 , 0, 3 × 10 kg · m/s
~F =
net,sys
3s
~F
D 3 3
E
net,sys
= −5 × 10 , 0, 1 × 10 N
2.X.8 This is another straightforward application of the momentum principle. Let the system consist of the particle and
assume non-relativistic speed.
∆~psys = ~Fnet,sys ∆t
~pf − ~pi = ~Fnet,sys ∆t
~pf = ~pi + ~Fnet,sys ∆t
~pf = h10, 0, 0i kg · m/s + (h−6, 3, 0i N) (0.1 s)
~pf ≈ h9.4, 0.3, 0i kg · m/s
, 3
2.X.9 Apply some straightforward reasoning rather than complicated mathematics. For the first hour of the trip, you’ll
travel 50 km. Neglect the duration of the change from 50 km/h to 100 km/h; either you hit the gas pedal hard (not good!) or
you have a car that accelerates quickly (preferable). During the next two hours of the trip, you’ll travel 200 km. You’ll travel
a total of 250 km during three hours. By definition, your average velocity will be vavg,x ≈ 2503km/h
h ≈ 83.3 km/h. This isn’t
equal to the arithmetic mean of the initial and final velocities because the force that caused the change in velocity wasn’t
constant for the three hour duration. This, in turn, means that the magnitude of the velocity could not change linearly.
NOTE: text has 50 km/h.
2.X.10
∆t = 10 s
~vi = h30, 0, 0i m/s
~vf = h40, 0, 0i m/s
Since ~v is changing at a constant rate,
vix − vf x
vavgx =
2
30 m/s + 40 m/s
=
2
= 35 m/s
∆x
vavgx =
∆t
∆x = vavgx ∆t
= (35 m/s)(10 s)
= 350 m
2.X.11
During each time interval, ∆px = 10 kg · m/s.
(a) False, because ∆px 6= 0
(b) True, because ∆px = 10 kg · m/s is constant, i.e. it is the same during each successive 1-second time interval.
(c) False, because ∆px is constant.
(d) False, because ∆px is constant.
2.X.12
2.X.1 Let ~pi = h1.5, −1.2, 0.7i kg · m/s and ~pf = h1.6, −0.9, 1.1i kg · m/s. The change in momentum is just ~pf - ~pi .
∆~p = ~pf − ~pi
= h1.6, −0.9, 1.1i kg · m/s − h1.5, −1.2, 0.7i kg · m/s
= h0.1, 0.3, 0.4i kg · m/s
2.X.2 The net force on a system is always the vector sum, or superposition, of the individual forces acting on the system.
~F = h40, −70, 0i N + h20, 10, 0i N
net,sys
= h60, −60, 0i N
2.X.3 This problem is best solved using simple ratio reasoning. Changing the force’s magnitude by a factor of two, for the
same duration, must also change the resulting change in momentum. Halving the force must halve the change in momentum.
So |~p| = 1.5 kg · m/s.
2.X.4 Let the object (whatever it is) be the system. We’re given that ~Fnet,sys = h−0.5, −0.2, 0.8i N and ∆t = 2 min = 120 s
(don’t forget to express ∆t in seconds!).
(a) We have to calculate the impulse as follows:
∆~p = ~Fnet,sys ∆t
= (h−0.5, −0.2, 0.8i N) (120 s)
= h−60, −24, 96i kg · m/s
Note that this is (must be!) the same value we would have gotten had we known both ~pf and ~pi .
(b) Remember that impulse and change in momentum are merely two different words for the same quantity. This quantity
can be called impulse if it is calculated from the net force and its duration. It can also be called change in momentum
if it is calculated from the initial and final momenta. Names aside, it is the same quantity in both cases!
2.X.5 Let the system consist of the hockey puck. Assume there are no significant interactions on the system other than from
ice and air (we can treat them as one net interaction), and this interaction was constant while it existed. We’re given that
~pi = h0, 2, 0i kg · m/s.
(a) If the puck comes to a stop, its final momentum must be zero. Therefore ~pf = h0, 0, 0i kg · m/s. The impulse, or change
in momentum, is merely ∆~p = ~pf - ~pi = h0, 0, 0i kg · m/s - h0, 2, 0i kg · m/s = h0, −2, 0i kg · m/s.
(b) We know ∆~p and ∆t so we can calculate ~Fnet,sys using the momentum principle.
∆~p = ~Fnet,sys ∆t
~F ∆~p
=
net,sys
∆t
~F h0, −2, 0i kg · m/s
=
net,sys
3
s
2
~F = 0, − , 0 N
net,sys
3
,2
2.X.6 This is a straightforward application of the momentum principle. Let the system be the puck, and assume the only
significant interaction on the system is with the hockey stick.
∆~p = ~Fnet,sys ∆t
~pf − ~pi = ~Fnet,sys ∆t
~pf = ~pi + ~Fnet,sys ∆t
~pf = h10, 0, 5i kg · m/s + h0, 0, 2000i N × (0.004 s)
~pf ≈ h10, 0, 13i kg · m/s
Note that the puck’s x and y components of momentum didn’t change because the net force had no components in those
directions.
2.X.7 We’re given that ~vi = h25, 0, 15i m/s, ~vf = h10, 0, 18i m/s, and mcar = 1000 kg. Let the system consist of the car,
and let’s assume non-relativistic speeds.
(a)
∆~psys = ~pf − ~pi
∆~psys ≈ mcar ~vf − mcar ~vi
∆~psys ≈ mcar ~vf − mcar ~vi
∆~psys ≈ (1000 kg) (h10, 0, 18i m/s − h25, 0, 15i m/s)
∆~psys ≈ (1000 kg) (h−15, 0, 3i m/s)
D 4 3
E
∆~psys ≈ −1.5 × 10 , 0, 3 × 10 kg · m/s
(b) Remember that impulse and change in momentum refer to the same physical quantity, so the answer is the same as
above.
(c)
∆~psys
~F =
D∆t
net,sys
4 3
E
−1.5 × 10 , 0, 3 × 10 kg · m/s
~F =
net,sys
3s
~F
D 3 3
E
net,sys
= −5 × 10 , 0, 1 × 10 N
2.X.8 This is another straightforward application of the momentum principle. Let the system consist of the particle and
assume non-relativistic speed.
∆~psys = ~Fnet,sys ∆t
~pf − ~pi = ~Fnet,sys ∆t
~pf = ~pi + ~Fnet,sys ∆t
~pf = h10, 0, 0i kg · m/s + (h−6, 3, 0i N) (0.1 s)
~pf ≈ h9.4, 0.3, 0i kg · m/s
, 3
2.X.9 Apply some straightforward reasoning rather than complicated mathematics. For the first hour of the trip, you’ll
travel 50 km. Neglect the duration of the change from 50 km/h to 100 km/h; either you hit the gas pedal hard (not good!) or
you have a car that accelerates quickly (preferable). During the next two hours of the trip, you’ll travel 200 km. You’ll travel
a total of 250 km during three hours. By definition, your average velocity will be vavg,x ≈ 2503km/h
h ≈ 83.3 km/h. This isn’t
equal to the arithmetic mean of the initial and final velocities because the force that caused the change in velocity wasn’t
constant for the three hour duration. This, in turn, means that the magnitude of the velocity could not change linearly.
NOTE: text has 50 km/h.
2.X.10
∆t = 10 s
~vi = h30, 0, 0i m/s
~vf = h40, 0, 0i m/s
Since ~v is changing at a constant rate,
vix − vf x
vavgx =
2
30 m/s + 40 m/s
=
2
= 35 m/s
∆x
vavgx =
∆t
∆x = vavgx ∆t
= (35 m/s)(10 s)
= 350 m
2.X.11
During each time interval, ∆px = 10 kg · m/s.
(a) False, because ∆px 6= 0
(b) True, because ∆px = 10 kg · m/s is constant, i.e. it is the same during each successive 1-second time interval.
(c) False, because ∆px is constant.
(d) False, because ∆px is constant.
2.X.12
