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MAT1501 EXAM PACK 2023

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  • October 13, 2023
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Assignment 2 Memo - Memo to help you check whether you
are on the right track or not
Fundamental Mathematics (University of South Africa)




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Downloaded by Jack martinez (pmachuki7686@gmail.com)

, lOMoARcPSD|24668432




SEMESTER 1

Assignment 2 [81]
Due date: 16 March 2020
unique number (Code):
Covered book chapters: 8, 9, 21, and 26

Questions

1. Find the gravitational potential energy of an 88-kg person standing atop
Mt. Everest at an altitude of 8848 m. Use sea level as the location for y = 0.

Solution (3)

U  m g y  ø88 kg ù ø 9.81 m/s 2 ù ø8848 m ù  7.6 106 J  7.6 MJ

2. A vertical spring stores 0.962 J in spring potential energy when a 3.5-kg
mass is suspended from it.

(a) By what multiplicative factor does the spring potential energy
change if the mass attached to the spring is doubled?

(b) Verify your answer to part (a) by calculating the spring potential
energy when a 7.0-kg mass is attached to the spring.

Solution (8)

(a) Doubling x will increase the potential energy by a factor of 4 because U is
proportional to x2

x1  m g k
(b) x  2m g k  2 x
2 1


U 2 12 k x22 ø 2 x1 ù
2

  4
U1 12 k x12 x12
U 2  4 U1  4 ø 0.962 J ù  3.85 J




Downloaded by Jack martinez (pmachuki7686@gmail.com)

, lOMoARcPSD|24668432




3. A 2.9-kg block slides with a speed of 2.1 m/s on a frictionless horizontal
surface until it encounters a spring.
(a) If the block compresses the spring 5.6 cm before coming to rest,
what is the force constant of the spring?
(b) What initial speed should the block have to compress the spring by
1.4 cm?

Solution (8)

Ki  U i  Kf  U f

(a)
1
2
m vi2  0  0  12 k xmax
2



m v 2 ø 2.9 kg ùø 2.1 m/s ù
2

k 2i   4080 N/m  4.1 kN/m
ø 0.056 m ù
2
xmax


ø 4100 N/m ùø 0.014 m ù
2 2
k xmax
(b) vi    0.53 m/s
m 2.9 kg


4. What is the mass of a mallard duck whose speed is 8.9m/s and whose
momentum has a magnitude of 11 kg.m/s?

Solution (3)


p 11 kg  m/s
p  mv  m    1.2 kg
v 8.9 m/s

5. A 26.2-kg dog is running northward at 2.70 m/s, while a 5.30-kg cat is
running eastward at 3.04 m/s. Their 74.0-kg owner has the same
momentum as the two pets taken together. Find the direction and
magnitude of the owner’s velocity.

Solution (12)
p total  p cat  p dog  mcat v cat  mdog v dog
 ø 5.30 kg ùø 3.04 m/s xˆ ù  ø 26.2 kg ùø 2.70 m/s yˆ ù
p total  ø16.1 kg  m/s ù xˆ  ø 70.7 kg  m/s ù yˆ




Downloaded by Jack martinez (pmachuki7686@gmail.com)

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