p10 009

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9.

(a) The initial momentum of the car is 

p

i

= m

v

i

= (1400 kg)(5.3 m/s)ˆj = (7400 kg

· m/s)ˆj and the final

momentum is 

p

f

= (7400 kg

· m/s)ˆi. The impulse on it equals the change in its momentum:



J = 

p

f

− p

i

= (7400 N

·s)(ˆiˆj) .

(b) The initial momentum of the car is 

p

i

= (7400 kg

· m/s)ˆi and the final momentum is p

f

= 0. The

impulse acting on it is



J = 

p

f

− p

i

=

7400ˆi N·s .

(c) The average force on the car is



F

avg

=



p

t

=



J

t

=

(7400 kg

· m/s)(ˆiˆj)

4.6 s

= (1600 N)(ˆi

ˆj)

and its magnitude is F

avg

= (1600 N)

2 = 2300 N.

(d) The average force is



F

avg

=



J

t

=

(

7400 kg · m/s)ˆi

350

× 10

3

s

= (

2.1 × 10

4

N)ˆi

and its magnitude is F

avg

= 2.1

× 10

4

N.

(e) The average force is given above in unit vector notation. Its x and y components have equal

magnitudes. The x component is positive and the y component is negative, so the force is 45

below the positive x axis.


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