AS and A Level Physics Original material © Cambridge University Press 2010 1

4 Marking scheme: Worksheet (AS) 1 D [1]

2 D [1]

3 C [1]

4 C [1]

5 A [1]

6 a R2 = 7.02 + 5.02 [1] R = 2549 + = 8.6 N [1]

b Force vertically = 40 − 10 = 30 N and force horizontally = 80 − 20 = 60 N [1] R2 = 302 + 602 [1]

N 673600900 ≈+=R [1]

7 a Fx = F cos θ = 10 cos 45° N 1.7N 07.7 ≈=xF [1]

Fy = F sin θ = 10 sin 45° N 1.7N 07.7 ≈=yF [1]

b Fx = F cos θ = 85 cos 20° N 80N 9.79 ≈=xF [1]

Fy = F sin θ = 85 sin 20° N 29N 1.29 ≈=yF [1]

8 a moment = force × perpendicular distance from the pivot to the line of action of the force [1] b The net force acting on the object is zero. [1] The sum of the clockwise moments about any point is equal to the sum of the

anticlockwise moments about the same point. [1]

9 a The centre of gravity of the beam is 0.75 m away from support A. sum of clockwise moments = sum of anticlockwise moments [1] 60 × 0.75 = RB × 1.5 [1] RB = 30 N [1]

b Net force in the vertical direction = 0 RA + RB = 60; RA = 60 − 30 = 30 N [1]

4 Marking scheme: Worksheet (AS)

AS and A Level Physics Original material © Cambridge University Press 2010 2

10 a The net force is zero because the seat is in equilibrium. [1]

b i

Correct diagram [1]

Weight = mg = 35 × 9.81 N 343≈ [1]

T2 = 1802 + 3432 [1]

N 390343180 22 ≈+=T [1]

ii tan θ = 343180 [1]

°≈= − 28)525.0(tan 1θ [1]

11 a Fx = F cos θ = 300 cos 30º [1] N 260≈xF [1]

b The net force is zero, because the roller is moving at constant velocity. [1] Resistive force N 260≈ to the left. [1]

c Fy = F sin θ = 300 sin 30° = 150 N [1]

The net vertical force is zero.

150 + contact force = mg [1] contact force = (50 × 9.81) − 150 N 340≈ [1]

12 a The line of action of the weight of the ladder is at a perpendicular distance of 0.75 m away from the foot of the ladder. Taking moments about the base of the ladder ⇒ sum of clockwise moments = sum of anticlockwise moments [1] (32 × 9.81) × 0.75 = R × 4.0 [1]

N 590.4

75.081.932≈

××=R [1]

b The force at the base of the ladder creates zero moment about this point. [1]

4 Marking scheme: Worksheet (AS)

AS and A Level Physics Original material © Cambridge University Press 2010 3

13 a

All forces clearly shown on the diagram. [2]

b Taking moments about the brick ⇒ sum of clockwise moments = sum of anticlockwise moments (62 × 9.81)x + (15 × 9.81) × 0.78 = (30 × 9.81) × 1.56 [2]

m 57.062

)1578.0()3056.1(≈

×−×=x [1]

m 57.0≈x Distance of centre of gravity from the toes = 1.56 − 0.57 = 0.99 m [1]