CIE 9702 Physics Paper 21 MJ 2015 Mark Scheme Answers

Q1(a)

Power = Work done / Time
Unit of power = kg m s^-2 m / s
                        = kg m² s^-3

Q1(b)

Power = p.d. x current
    p.d. = power / current

unit of p.d. = kg m² s^-3/ A
                   = kg m² s^-3 A^-1

Q2(a)

speed: distance travelled per unit time. It is a scalar quantity because distance has only magnitude
velocity: displacement per unit time. It is a vector quantity because displacement has magnitude and direction

Q2(b)

(i) The velocity increases at a constant rate from 0 m s^-1 to 11.0 m s^-1, it decreases and changes direction in 0.1 s second upon impact with the ground. After rebound, the velocity decreases to zero at the same rate as initial

(ii) a = v - u / t
         = (8.8 + 8.8) / 1.8
         = 9.8 m s^-2

(iii)
1. distance = first area above graph + second area below graph
                  = (1.1 × 10.8) / 2 + (0.9 × 8.8) / 2 (= 5.94 + 3.96)
                  = 9.9 m

2.displacement = first area above graph – second area below graph
                          = (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2
                          = 2.0 m

(iv)

Flip the negative portion of the original graph to positive
Note: Since question asked for 'speed', all values should be positive.

Q3(a)

MxUx + MyUy = MxVx + MyVy
(50)(+4.5) + (My)(-2.8) = 50(-1.8) + (My)(+1.4)
My = 75 g

Q3(b)

Collision is not elastic.
because the total initial KE NOT equal to total final KE.

Note: You need to specify that it is the 'total' KE, a brief sentence of 'kinetic energy is not conserved' is not sufficient.
Also, show your calculation within the space given.

Q3(c)

According to N3L, force on X is equal and opposite to force on Y
According to N2L, force is proportional to rate of change of momentum,
since time of collision same for both balls, their change in momentum is the same.

Note: Apply the laws specifically to this situation, not a mere statement of the two laws.

Q4(a)

(i) gradient of the graph is (4.5 - 1.5) / (1.8 - 0.6) = 2.5
                 y = mx + c, 4.5 = 2.5(1.8) + c, c = 0
      so it is a straight line passing through origin,
      hence compression of the spring obeys Hooke's law

(ii)  k = F / x = 4.5 / 0.018 = 250 N m^-1

(iii) PE = 1/2 F x = area under the graph = 1/2 (4.5) (0.018) = 0.041 J
Note: Area under the graph is Not 1/2(4.5-1.5)(0.018-0.006), as compression should start at 0 cm

Q4(b)

KE = PE
1/2 m v² = 0.041
            v = 0.22 m s^-1

Q5(a)

From -0.5 V to 0.45 V, resistance is very large. After 0.45 V, resistance decreases quickly and reaches a very low value within 0.35V.

Q5(b)

Initial straight line from (0,0) into curve with a decreasing gradient, but not achieving horizontal Repeat in the negative quadrant.

Q5(c)

(i) When the p.d. is 12 V across the lamp, it will have a power of 36 W,
     P = V² / R
    36 = 12² / R
     R = 4.0 ohm

(ii) emf = current x total resistance
          E = I (R +r)
         12 = 2.8 (R + 0.50)
           R = 3.79 ohm

Q5(d)

When the p.d. across the filament lamp decreases, the resistance decrease.

Q6(a)

diffraction: the spreading of the wave as it passes through a slit or past an edge
interference: when two or more waves superpose/overlap/meet at a point, the resultant displacement is the sum of the displacement of each wave.
Note: the description for interference sounds more like the principle of superposition, but it shall also be the way of how the phenomena of interference is explained

Q6(b)

nλ = d sin θ and v = fλ
max order number can be calculated by setting θ = 90°
hence n = d / λ
             = (1/N) (f / v)
             = [ 1 / (650 × 10³) ] (7.06 × 10¹⁴ / 3 × 10⁸ )
          n = 3.6
hence number of orders = 3

Q6(c)

Lower frequency means bigger wavelength.
Based on the equation derived above, n = d / λ,
bigger wavelength will give a smaller n, so fewer orders will be seen.

Q7(a)

Electric field is a region where a charge experiences an electrical force.

Q7(b)

(i) Draw at least four parallel equally spaced straight lines perpendicular to plates,
     direction arrows from left (AB) to right(CD)

(ii) E = V / d
          = 450 / 16 x 10^-3
          = 28 x 10³

(iii) W = q V
            = 2 x 1.6 x10^-19 x 450
            = 1.4 x 10^-16

(iv) W = q V, W directly proportional to q when V is the same.
       Charge of alpha particle is two times of beta particle
       So, work done on alpha particle is two times of beta particle
       Walpha / Wbeta = [ 2 x 1.6 x10^-19 x 450 ] / [ 1.6 x10^-19 x 450 ]
                                  = 2