A-LevelPhysicsKinematics9702/22 · Feb/Mar 2025 · Q310 Marks

A truck R of mass 9400kg moves with constant acceleration in a straight line down a slope, as illustrated in [Figure 3.1]. At point A the speed of the truck is 13ms⁻¹ and at point B the speed of the truck is 22ms⁻¹. A and B are a distance of 180 m apart.

(a)A truck R of mass 9400kg moves with constant acceleration in a straight line down a slope, as illustrated in [Figure 3.1]. At point A the speed of the truck is 13ms⁻¹ and at point B the speed of the truck is 22ms⁻¹. A and B are a distance of 180 m apart.
(a)(i)Calculate the acceleration of the truck between A and B. acceleration = ................................................................ms⁻² [2][2]
(a)(ii)Determine the gain in kinetic energy of the truck between A and B. gain in kinetic energy = ................................................................J [3][3]
(b)A short time after passing point B truck R moves in a straight line on horizontal ground. The driver of the truck applies the brakes. [Figure 3.2] shows the variation with time of the momentum of the truck.
(b)(i)Define force.[1]
(b)(ii)Show that the average resultant force F acting on truck R between time t = 0 and t = 15s is -1.2 × 10⁴ N.[1]
(b)(iii)An identical truck S has the same initial momentum as truck R. Truck S experiences a constant force equal to the force F in (b)(ii). State and explain whether truck S will take more, less or the same amount of time to come to rest as truck R.[3]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

C1 a = (v² – u²) / 2s
= (222 – 132) / (2 × 180)
A1 = 0.88 m s-2
OR
[t = (180 × 2) / (22 + 13) = 10.3]
180 = 13 x 10.3 + 1/2 a x 10.32 or
180 = 22 x 10.3 - 1/2 a x 10.3 2 or
22 = 13 + a x 10.3
(C1)
(A1) a = 0.88 m s-2
C1 (Δ)E = 1/2m(∆)v²
C1 gain in KE = 1/2m(v² – u²)
= 1/2 × 9400 × (222 – 132)
A1 = 1.5 × 106 J
OR
W = Fs
= ma x d
(C1)
gain in KE = 9400 × 0.88 × 180
(C1)
(A1) = 1.5 × 106 J
B1 rate of change of momentum
A1 (Force =) (2.5 × 104 – 21 × 104) / 15 = – 1.2 × 104 (N)
B1 The change of momentum (for S and R to come to rest) is the same
B1 (Average) force on R (to come to rest) is (-21 × 104/23 =) 0.91 × 104 N or (Average) force on R (to come to rest) is less than the force on S / less than F
B1 (S will come to rest in) less time (than R).
OR
(B1) The change of momentum (for S and R to come to rest) is the same
(B1) Time for S to come to rest is (−21 × 104 / 1.2 × 104 =) 17.5 s (and time for R to come to rest is 23 s)
(B1) (S comes to rest in) less time (than R).

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