# 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 v

^{2}= u^{2}+ 2as,}
v

^{2}= 0 + 2(9.81)(25) (or using ½mv^{2}= 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 + ½at

^{2}where u = 0}
22.1 = 0 + 9.81t (or 25 = ½ ×
9.81 × t

^{2})
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}= 25^{2}+ 33.86^{2}
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 mv

^{2}D mv^{2}**Reference:**

*Past Exam Paper – June 2015 Paper 11 Q19*

__Solution 1017:__**Answer: D.**

This was a difficult question.

Sand drops vertically on to the belt
at a rate of m kg s

^{–1}. [Î”mass / time = m kg s^{–1}]
Consider momentum. Momentum p = mass
× velocity

Rate of change of momentum = Î”p /
t = (Î”mass × velocity) / t

Rate of change of momentum = (Î”mass / t) × velocity) = mv {note that here m is not the mass, but the rate of drop of
the sand}

In words, the above equation for the
rate of change of momentum is be explained as:

In one second, the momentum of mass
m of sand increases from zero to mv.

Remember that force is defined as
the rate of change of momentum. So the force involved is

**mv**and the power (= Force × velocity) required is mv^{2}.
Note that the kinetic energy of the
sand does increase by ½ mv

^{2}but this cannot be the only power involved (it would imply an infinite acceleration for every grain of sand landing on the belt).

__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:__**(a)**

(i) Vertical component = 44 sin 30° = 22 N

(ii) Horizontal component = 44 cos 30° = 38(.1) N

**(b)**

{The vertical component of F acts at a distance of
(0.64 + 0.96 =) 1.60m from the pivot.

Clockwise moment = Anti-clockwise moment}

W × 0.64 = 22 × 1.60

W = 55 N

**(c)**F has a horizontal component (not balanced by W)

{The horizontal component of F acts on the wall and
this is not balanced by W. From Newton’s 3

^{rd}law, there should be an equal and opposite force exerted by the wall on the rod.}**(d)**Line from P in the direction towards the point on wire vertically above W and direction up

{This force, along with F and W should form a system
that is in equilibrium. For equilibrium, the resultant force and the resultant
moment should be zero. If all the 3 forces pass through the same point, the
resultant moment would be zero.

Moment = Force × perpendicular distance from line
of action of force to pivot.

In this case, the pivot is that point where all the 3
forces pass. As they pass on the point, the ‘distance …’ is zero, and thus, the
moment at that point is zero.}

__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 mv

^{2}
KE = 0.5 × 0.4 × (15.0)

^{2}= 45 J**(c)**

{Equation for uniformly accelerated motion: s = ut + ½
at

^{2}.
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 at

^{2}= 0 + ½ × 2.5 × t^{2}
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

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