Download Question 1 - RobboPhysics

UNIT 4.
PART D: STRETCHED SPRINGS
SOLUTIONS
Question 1
K E = 0. 5 X 0. 6 5 x (7.1)2.' = 16.6 Joules.
Each square under the graph has area of 0.005 X 100 = 0.5 Joules.
Number of squares = 31. "0.040" gives 32 squares. 0.040 m (C).
Alternatively, calculate the ‘k' value use Area = 0.5 kx2.
Question 2
Gain in KE = Work Done = Force x Distance. Estimate the distance the hand and ball move during the
throw: approx 0.5 - 1 metre. Force is approximately 24 - 12 Newton.
Question 3
Wtotal = 4 Wspring = 4 kx2/2 = 2 x 2.5 x 104 (0.04)2 = 80 j
Question 4
60%
Question 5
½ kx2 = 200J
0.10m
Question 6
GPE: Tyre => Compress spring
mgh = 50 x 10 x 1 = 500 J
300 J converted to heat / vibrational motion – spring / tyre heat up.
Question 7
76%
Original KE = work done in compressing air bag = area under graph = 800 + 6000 = 6800 J
Final KE = work done in expansion = 1600 J
Fraction dissipated = 5200/6800 = 0.7647
Question 8
In an impact the bead's momentum is reduced to zero and transferred to the earth. Its kinetic energy is
dissipated. The impulse, which is equal to the change in momentum, and the work done, which is
equal to the change in KE, will be the same regardless of the nature of the impact. However, a helmet
enables the time of impact to be extended, thus reducing the force as the impulse is unchanged.
Similarly, the loss of energy occurs over a greater distance, which also means less force.
Question 9
The head is attached to a trunk which may be restrained by a seatbelt.
Question 10
Grav PE = KE, mgh = ½ mv2.
5.0 x 10 x 1.5 = ½ x 5.0 x v2
v = 5.48 m/s
Question 11
Area under graph = ½ x 30000 x 0.2 = 300 J
Question 12
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UNIT 4.
PART D: STRETCHED SPRINGS
SOLUTIONS
Find Force when-area = 75 Joules.
75 = ½ kx2, where k = gradient = 1.5 x 106 N/m
75 = ½ (1.5 x 106 )x2.
x = 1.0 x 10-2
Force = 1.5 x 104 N.
Alternatively 75 = 1/4 x 300 joules = ¼ x area
Force = 1.0 x 104 N.
Question 13
Force mass x accel'n
Accel'n = 2.0 x 104 N / 5.0 kg = 4.0 x 103 m/s2
Question 14
For Force = max value of 22.0 x 104 N, area = 50 joules of energy is stored, so therefore if 75 joules is
to
be stored, then the force must be greater, so the helmet does not fit the design standard.
Question 15
Elastic behaviour up to 0.4 m extension. Strain energy is area under graph to this point.
Area = ½ x 0.4 x 40 = 8 J
Question 16
When extension is 0.3 m, force = 30 N
mass = 5.0 kg
F
 6 m/s2
a=
M
Question 17
Kinetic energy gained = area under graph to 0.3 m
= ½ x 0.3 x 30 = 4.4 J
½ mv2 = 4.5 J  v = 1.3 m/s
Question 18
a)The weight force of the car will be shared equally between the 4 springs
each spring supports 1000/4 = 250 kg =2.5 x 103 N
2.5 x10 3
From Hooke’s law: force constant = k = F/x =
= 2.5 x 105 Nm-1
1x10 2
b)average elastic potential energy stored in each spring
Epelastic = ½ kx2 = ½ x 2.5 x 105 x (0.01)2 = 12.5 J
Question 19
Torques about pivot are equal 200 x 40 - 40 x 30= 8000 - 1200 = F x 17. F = 6800/17 = 400N
Question 20
F = kx. 40 N = k x 0.06. k = 40 / 0.06 = 670 N/m
Question 21
36 kph = I0ms-1
Velocity Change = -10 ms-1 Area under graph = 10 ms-1
0.5 x 40 x 500 x 10-3 =
10 ms-1
Collision time = 40 x 10-3 s Collision time =0.04 s (40 ms)
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UNIT 4.
PART D: STRETCHED SPRINGS
SOLUTIONS
alternatively:
area of one square of graph = 1 ms-1
need 10 squares
 time ~ 40 ms
Question 22
Av. Force = Av. acceleration x rnass = 250 x 1000 = 2.5 x 105 N
alternatively:
mv 1000 x10

F=
= 2.5 x 105 N
3
t
40 x10
Question 23
K.E. = 0.5 x 1000 x 102
= 5.0 x 104 J
K.E.loss = Av. Force x Distance
5.0 x10 4
Distance =
m
2.0 x 10-1 m = 20 cm
2.5 x10 5
Question 24
The velocity is initially positive and goes to zero. The decrease in velocity is slow to start then
increases.
Therefore the acceleration graph has a gradient starting at zero becoming increasingly negative => F
Question 25
Energy = area under graph = 0.5 x 0.30 x 6000 = 900 J
Question 26
KE at floor = GPE at height = 0.25 x 10 x 3.0 = 7.5 joules
Question 27
Stored elastic PE = 0.5kx2 = 0.5 x 3.40 x 104 x (0.020)2 = 6.8 joules
Question 28
Energy lost has gone into heat, noise, which is KE of atoms and molecules, and permanent deformation
or damage of the floor and ball.
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