Saturday, October 17, 2015

Physics 9702 Doubts | Help Page 212

  • 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:
Answer: D.
This was a difficult question.

Sand drops vertically on to the belt at a rate of m kg s–1. [Δmass / time = m kg s–1]
Consider momentum. Momentum p = mass × velocity
Rate of change of momentum = Δp / t = (Δmass × velocity) / t
Rate of change of momentum = (Δmass / 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 mv2.


Note that the kinetic energy of the sand does increase by ½ mv2 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 3rd 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 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.}



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