A-LevelPhysicsForces, density and pressure9702/23 · May/June 2017 · Q19 Marks

Two forces, with magnitudes 5.0N and 12N, act from the same point on an object. Calculate the magnitude of the resultant force R for the forces acting An object X rests on a smooth horizontal surface. Two horizontal forces act on X as shown in [Figure 1.1]. A force of 55N is applied to the right. A force of 18N is applied at an angle of 115° to the direction of the 55N force.

(a)(i)in opposite directions,[1]
(a)(ii)at right angles to each other.[1]
(b)(i)Use the resolution of forces or a scale diagram to show that the magnitude of the resultant force acting on X is 65 N.[2]
(b)(ii)Determine the angle between the resultant force and the 55N force.[2]
(c)A third force of 80 N is now applied to X in the opposite direction to the resultant force in (b). The mass of X is 2.7 kg. Calculate the magnitude of the acceleration of X.[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

1(a)(i) R = 7(.0) N
1(a)(ii) R = 13 N
1(b)(i) forces resolved: 18 sin 65° (vertical) and 55 + 18 cos 65° (horizontal)
1(b)(i) or scale drawing: correct triangle drawn for forces
1(b)(i) F = [(18 sin 65°)² + (55 + 18 cos 65°)²]1/2 = 65 (64.7) N
1(b)(i) or scale drawing: scale given, length of resultant given correctly, ± 1 N
1(b)(ii) angle = tan¯¹ [18 sin 65° / (55 + 18 cos 65°)] = tan¯¹ (16.3/62.6)
1(b)(ii) or scale drawing: correct angle measured/direction correct on diagram below the 55 N force
1(b)(ii) angle = 15 (14.6)° (below the 55 N force)
1(b)(ii) or scale drawing: angle = 15° ± 1°
1(c) (resultant) force = mass x acceleration
1(c) 80-65 = 2.7a
1(c) a = 5.6 m s¯² [5.7 if 64.7 N used from (i)]

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