FND sample solutions
PHYSICS 1002
Mid-Semester Test SAMPLE SOLUTIONS
OIL ANDWATER
Q7. The pressure at level C is given by the sum of the pressure of the water, plus the pressure of the
oil, plus the presure of the atmosphere. If the total depth is h, then the pressure is
pC = p0 + ρwg
(
h
2
)
+ ρoilg
(
h
2
)
(equation not required)
(2 marks)
Q8. From Q7, the pressure at C is
pC = 1.013× 105 + (1.00× 103)(9.8)(0.23) + (0.850× 103)(9.8)(0.23)
= 1.013× 105 + 2.25× 103 + 1.92× 103 = 1.055× 105 Pa
(3 marks)
Q9. The cube comes to rest at the interface. The density of the wood is greater than the density of oil,
so it sinks in oil; but the density is less than the density of water, so it floats in water. Hence it
comes to rest at the interface, partly submerged in the water.
(3 marks)
FND sample solutions Semester 1 Page 2 of 3
SKYDIVERS
Q10. The first skydiver has initial velocity v0 = 0 m/s, and acceleration a = −g = −9.81 m s−2
vertically downward.
(2 marks)
Q11. Yes, the difference in the velocities of the skydivers stays the same. If we start the clock when
skydiver 2 jumps out, then skydiver 1 has a greater initial velocity. Since both are accelerating
due to gravity, then each second their velocity increases by the same amount, so the difference
between their velocities remains the same.
Or: version using equations:
Yes, the difference in the velocities of the skydivers stays the same. Start the clock when skydiver
2 jumps: so skydiver 1 jumps at t = −∆t. At a later time t, his velocity is
v1 = 0 + a(t+ ∆t) = at+ a∆t = v2 + a∆t
where v2 = at is the velocity of skydiver 2. So v1 − v2 = a∆t = constant.
(3 marks)
Q12. Because there is a velocity difference between them, then they will move apart: in t seconds,
skydiver 1 travels a distance v1t, while skydiver 2 travels a distance v2t, which is > v1t since
v2 > v1. This means the elastic cord will stretch and therefore there will be an increase in
tension.
(3 marks)
FND sample solutions Semester 1 Page 3 of 3
MONKEY AND ZOOKEEPER
Q13.
vix = 60 m s−1 cos(36.9◦) = 48.0 m s−1
(1 mark)
Q14.
viy = 60 m s−1 sin(36.9◦) = 36.0 m s−1
(1 mark)
Q15.
t =
12.0
48.0
= 0.25 s
(1 mark)
Q16. Use d = ut+ 12 t
2 with u = viy = 36.0 m s−1, t = 0.25 s and a = −9.81 m s−2:
d = 36.0 × 0.25 + 0.5 × (−9.81)× (0.25)2 = 8.7 m
(2 marks)
Q17. Use d = ut+ 12 t
2 with u = 0 m s−1, t = 0.25 s and a = −9.81 m s−2:
d = 0 + 0.5 × (−9.81)× (0.25)2 = −0.3 m
(1 mark)
Q18. Yes, the monkey is hit by the dart. Because both the dart and the monkey are accelerating due to
gravity at the same rate, the distance the dart has fallen from where it was aimed (9.0 − 8.7 =
0.3 m) is the same as the distance the monkey has fallen from rest (0.3 m). So dart and monkey
end up at the same height, so the dart hits the monkey.
(2 marks)
THERE ARE NOMORE QUESTIONS.