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6677 edexcel gce mechanics m1 advanced advanced subsidiary 4

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1. A 3 kg object is attached to a spring with a spring constant of 200 N/m. What is the extension
in the spring when the object is at rest?

A. 0.1 m

B. 0.15 m

C. 0.2 m
D. 0.25 m

Answer: A) 0.1 m

Rationale: Using Hooke’s law, F=kxF = kxF=kx, where F=mg=3×9.8=29.4F = mg = 3 \times 9.8
= 29.4F=mg=3×9.8=29.4 N, and k=200k = 200k=200 N/m. So, x=Fk=29.4200=0.147≈0.1x =
\frac{F}{k} = \frac{29.4}{200} = 0.147 \approx 0.1x=kF=20029.4=0.147≈0.1 m.



2. A car is moving at a constant speed of 20 m/s. What is the car’s kinetic energy?

A. 80,000 J

B. 100,000 J

C. 120,000 J

D. 160,000 J

Answer: B) 100,000 J
Rationale: KE=12mv2KE = \frac{1}{2} mv^2KE=21mv2,
KE=12×1000×202=100,000KE = \frac{1}{2} \times 1000 \times 20^2 =
100,000KE=21×1000×202=100,000 J.



3. A particle of mass 5 kg is moving with a velocity of 2 m/s. What is its momentum?

A. 2 kg·m/s

B. 5 kg·m/s

C. 10 kg·m/s

D. 20 kg·m/s

,Answer: C) 10 kg·m/s

Rationale: Momentum = mass × velocity. So, 5×2=105 \times 2 = 105×2=10 kg·m/s.



4. A stone is dropped from the top of a building. After 3 seconds, what is its velocity?

A. 9.8 m/s

B. 19.6 m/s

C. 29.4 m/s
D. 39.2 m/s

Answer: B) 29.4 m/s

Rationale: Using v=u+gtv = u + gtv=u+gt, where u=0u = 0u=0 (initial velocity), g=9.8g =
9.8g=9.8 m/s², and t=3t = 3t=3 s, we have v=0+9.8×3=29.4v = 0 + 9.8 \times 3 =
29.4v=0+9.8×3=29.4 m/s.



5. A force of 20 N is applied to a 4 kg object. What is the acceleration?

A. 2 m/s²

B. 4 m/s²

C. 5 m/s²

D. 8 m/s²

Answer: B) 5 m/s²
Rationale: Using F=maF = maF=ma,
a=Fm=204=5a = \frac{F}{m} = \frac{20}{4} = 5a=mF=420=5 m/s².



6. A particle moves in a circle with radius 5 m. If the particle has a speed of 10 m/s, what is its
centripetal acceleration?

A. 2 m/s²

B. 5 m/s²

C. 10 m/s²

D. 20 m/s²

, Answer: C) 10 m/s²

Rationale: Centripetal acceleration is given by ac=v2ra_c = \frac{v^2}{r}ac=rv2,

ac=1025=20a_c = \frac{10^2}{5} = 20ac=5102=20 m/s².



7. A car with a mass of 1000 kg is moving at 15 m/s. What is its kinetic energy?

A. 112,500 J

B. 150,000 J
C. 225,000 J

D. 250,000 J

Answer: A) 112,500 J
Rationale: KE=12mv2KE = \frac{1}{2} mv^2KE=21mv2,

KE=12×1000×152=112,500KE = \frac{1}{2} \times 1000 \times 15^2 =
112,500KE=21×1000×152=112,500 J.



8. A mass of 2 kg is at rest on a horizontal surface. What is the frictional force acting on the
object if there is no applied force?

A. 0 N

B. 2 N

C. 10 N
D. 20 N
Answer: A) 0 N

Rationale: If no force is applied, the frictional force is zero.



9. A block of mass 10 kg is pulled up a 30° incline. If the applied force is 50 N, what is the
acceleration?

A. 2 m/s²

B. 4 m/s²

C. 6 m/s²
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