Physics 9702 Doubts | Help Page 212
Question 1016: [Kinematics > Projectile motion]
(a) Explain what is meant by a scalar quantity and by a vector
quantity.
(b) A ball leaves point P at the top of a cliff with a horizontal
velocity of 15 m s–1, as shown in Fig. 2.1.
The height of the cliff is 25 m. The
ball hits the ground at point Q.
Air resistance is negligible.
(i) Calculate the vertical velocity
of the ball just before it makes impact with the ground at Q.
(ii) Show that the time taken for
the ball to fall to the ground is 2.3 s.
(iii) Calculate the magnitude of the
displacement of the ball at point Q from point P.
(iv) Explain why the distance
travelled by the ball is different from the magnitude of the displacement of
the ball.
Reference: Past Exam Paper – June 2014 Paper 23 Q2
Solution 1016:
(a)
A scalar quantity has a magnitude
only.
A vector quantity has both a
magnitude and a direction.
(b)
(i)
{initial vertical velocity
is zero. Using v2 = u2 + 2as,}
v2 = 0 + 2(9.81)(25) (or using ½mv2 = mgh {considering vertical motion only})
Vertical speed at Q, v = 22(.1) ms-1
{Note that this is the
vertical component of the velocity, not the resultant velocity since the ball
also has a horizontal component.}
(ii)
{Using v = u + at. For the
OR case, s = ut + ½at2 where u = 0}
22.1 = 0 + 9.81t (or 25 = ½ ×
9.81 × t2)
Time taken, t (= 22.1 / 9.81) =
2.26s or t [= √(5.097)]
= 2.26s
(iii)
Horizontal distance travelled = 15 × t =
15 ×
2.257 = 33.86 (allow 15×2.3
= 34.5)
{From Pythagoras’
theorem,}
(displacement)2 = (horizontal
distance)2 + (vertical distance)2
(displacement)2 = 252
+ 33.862
Magnitude of displacement = 42m
(42.08)
(iv) The distance travelled is the
actual (curved) path followed by the ball while the displacement is the
straight line / minimum distance from P to Q
Question 1017:
[Work, Energy and Power + Momentum]
A conveyor belt is driven at
velocity v by a motor. Sand drops vertically on to the belt at a rate of m kg s–1.
What is the additional power needed
to keep the conveyor belt moving at a steady speed when the sand starts to fall
on it?
A ½ mv B mv C ½ mv2 D mv2
Reference: Past Exam Paper – June 2015 Paper 11 Q19
Solution 1017:
Go to
Question 1018: [Dynamics > Moments > Equilibrium]
A rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1.
The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m
from P. The rod is kept horizontal and in equilibrium by a wire attached to Q
and to the wall at R. The wire provides a force F on the rod of 44 N at 30° to
the horizontal.
(a) Determine
(i) the vertical component of F,
(ii) the horizontal component of F.
(b) By taking moments about P, determine the weight W of the rod.
(c) Explain why the wall must exert a force on the rod at P.
(d) On Fig. 3.1, draw an arrow to represent the force acting on the rod at
P. Label your arrow with the letter S.
Reference: Past Exam Paper – June 2015 Paper 22 Q3
Solution 1018:
Go toA rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1. The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m from P.
Question 1019: [Kinematics > Linear motion]
The variation with time t of the velocity v of a ball is shown in Fig.
2.1.
The ball moves in a straight line from a point P at t = 0. The mass of
the ball is 400 g.
(a) Use Fig. 2.1 to describe, without calculation, the velocity of the ball
from t = 0 to t = 16 s.
(b) Use Fig. 2.1 to calculate, for the ball,
(i) the displacement from P at t = 10 s,
(ii) the acceleration at t = 10 s,
(iii) the maximum kinetic energy.
(c) Use your answers in (b)(i) and (b)(ii) to determine the time from t = 0
for the ball to return to P.
Reference: Past Exam Paper – June 2015 Paper 23 Q2
Solution 1019:
(a) There is a constant rate of increase in velocity/acceleration
from t = 0 to t = 8 s. There is a constant deceleration from t = 8 s to
t = 16 s OR constant rate of increase
in velocity in the opposite direction from t = 10 s to t = 16 s
(b)
(i)
Displacement = area under lines to 10 s
Displacement = {(5.0 × 8.0) / 2} + {(5.0 × 2.0) / 2} = 25 m
OR Displacement = ½ (10.0 × 5.0) = 25 m
(ii)
Acceleration a = (v – u) / t OR
gradient of line
{The velocity changes from +5 to -15 in a time of (16 –
8) seconds.}
Acceleration a = (–15.0 –5.0) / 8.0
Acceleration a = (–) 2.5 m s–2
(iii)
{The maximum KE depends on the velocity where the
magnitude is highest.}
KE = ½ mv2
KE = 0.5 × 0.4 × (15.0)2 = 45 J
(c)
{Equation for uniformly accelerated motion: s = ut + ½
at2.
The distance s moved from time t = 0 until the ball starts
moving towards P again is 25m. So, the ball should again travel a distance of
25m to return to P.
We now need to consider the motion during the
deceleration. Initial velocity u = 0. Deceleration a = 2.5ms-2. t is
the time from the instant it starts moving towards P, until it reaches P.}
distance = 25 = ut + ½ at2 = 0 + ½ × 2.5 × t2
Time t = 4.5 (4.47) s, therefore
time to return = {10 + 4.5 =} 14.5 s
{The backwards motion starts at t = 10s, so we need to
include these 10s too.}
tks
ReplyDeleteIts really healpfull. Tysm
ReplyDeletecould you explain the last line in the conveyor belt q
ReplyDeletecan you please explain the june 15 paper Q19 what do you mean by saying every grain would have an infinite acceleration?
ReplyDeletethe last line is not important to the explanation. it has been removed to prevent any confusion.
Deletethe explanation has also been updated.
October/November 2015 Paper 21 Question 4
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Q1018 are part c and d different? Are they talking about differen to forces
ReplyDeleteNo, they are the same force
DeleteMay/June 2017 22 question 1
ReplyDeletego to
Deletehttp://physics-ref.blogspot.com/2019/08/9702-june-2017-paper-22-worked-solution.html