Free practice final
Physics 1 practice final (free, with answers)
A typical Physics 1 final covers motion in one and two dimensions, Newton's laws with friction and circular motion, work and energy, momentum and collisions, rotation, and simple harmonic motion. Every problem here can be solved with algebra and trigonometry, so the page suits both algebra-based and calculus-based courses. Draw a diagram and work each problem on paper before opening the answer.
Topics
- Kinematics
- Projectile Motion
- Newton's Laws
- Friction and Inclines
- Circular Motion
- Work and Energy
- Momentum and Collisions
- Rotation and Torque
- Simple Harmonic Motion
15 questions
Try each one before you open its answer.
1. A car traveling at 30 m/s brakes with a constant deceleration of 6.0 m/s². How far does it travel before it stops?
- A 46 m
- B 75 m
- C 150 m
- D 225 m
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Answer: B. 75 m
With final speed zero, v² = v₀² − 2ad gives d = v₀²/(2a) = (30)²/(2 × 6.0) = 75 m. Forgetting the factor of 2 gives the wrong 150 m.
2. A ball is thrown straight up at 12 m/s. Ignoring air resistance and using g = 9.8 m/s², how high above the launch point does it rise?
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Answer: About 7.3 m
At the highest point the velocity is zero, so 0 = v₀² − 2gh gives h = v₀²/(2g) = (12)²/(2 × 9.8) ≈ 7.3 m.
3. A ball rolls off the edge of a 20 m high cliff with a horizontal speed of 12 m/s. Ignoring air resistance and using g = 9.8 m/s², how far from the base of the cliff does it land?
- A 15 m
- B 17 m
- C 24 m
- D 49 m
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Answer: C. 24 m
The fall time depends only on the vertical motion: 20 = ½(9.8)t² gives t = √(40/9.8) ≈ 2.02 s. The horizontal speed stays 12 m/s the whole way down, so the ball lands 12 × 2.02 ≈ 24 m from the base.
4. A 60 kg person stands on a bathroom scale in an elevator that is accelerating upward at 2.0 m/s². Using g = 9.8 m/s², what does the scale read?
- A 120 N
- B 468 N
- C 588 N
- D 708 N
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Answer: D. 708 N
Taking up as positive, Newton's second law gives N − mg = ma, so N = m(g + a) = 60(9.8 + 2.0) = 708 N. The scale reads more than the person's 588 N weight because the floor must also supply the upward acceleration.
5. Blocks of mass 2.0 kg and 3.0 kg rest on a frictionless horizontal surface, joined by a light string. A 20 N horizontal force pulls the 3.0 kg block, which drags the 2.0 kg block behind it. What is the tension in the string?
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Answer: 8.0 N
Treating both blocks as one system, a = 20 N / 5.0 kg = 4.0 m/s². The string is the only horizontal force on the 2.0 kg block, so T = (2.0 kg)(4.0 m/s²) = 8.0 N.
6. A 4.0 kg block slides down a 30° incline, and the coefficient of kinetic friction between the block and the incline is 0.20. Using g = 9.8 m/s², what is the magnitude of the block's acceleration?
- A 3.2 m/s²
- B 3.9 m/s²
- C 4.9 m/s²
- D 6.6 m/s²
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Answer: A. 3.2 m/s²
Along the incline, gravity pulls down the slope with mg sin 30° and kinetic friction acts up the slope with μmg cos 30°, so a = g(sin 30° − 0.20 cos 30°) = 9.8(0.500 − 0.173) ≈ 3.2 m/s². The mass cancels, and using sin 30° in the friction term instead of cos 30° gives the wrong 3.9 m/s².
7. A block rests on a board whose angle is slowly raised. The block starts to slide when the board makes an angle of 25° with the horizontal. What is the coefficient of static friction between the block and the board?
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Answer: μs = tan 25° ≈ 0.47
Just before slipping, static friction is at its maximum μs·N, and balancing forces along and perpendicular to the board gives mg sin θ = μs·mg cos θ. So μs = tan θ = tan 25° ≈ 0.47.
8. A car rounds a flat, unbanked curve of radius 50 m, and the coefficient of static friction between the tires and the road is 0.60. Using g = 9.8 m/s², what is the greatest speed at which the car can take the curve without skidding?
- A 5.5 m/s
- B 17 m/s
- C 22 m/s
- D 29 m/s
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Answer: B. 17 m/s
Static friction supplies the centripetal force, so at the limit μs·mg = mv²/r and v = √(μs·g·r) = √(0.60 × 9.8 × 50) = √294 ≈ 17 m/s. The mass cancels, so the speed limit does not depend on how heavy the car is.
9. A spring with k = 200 N/m is compressed 0.10 m and used to launch a 0.050 kg ball along a frictionless horizontal track. What is the ball's speed after it leaves the spring?
- A 4.5 m/s
- B 20 m/s
- C 28 m/s
- D 6.3 m/s
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Answer: D. 6.3 m/s
The spring's stored energy becomes kinetic energy: ½kx² = ½mv², so v = x√(k/m) = 0.10 × √(200/0.050) ≈ 6.3 m/s. Forgetting to square the compression gives the wrong 20 m/s.
10. A 2.0 kg box sliding across a level floor at 6.0 m/s comes to rest after sliding 4.0 m. Assuming the friction force is constant, what is its magnitude?
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Answer: 9.0 N
By the work-energy theorem, the work done by friction equals the change in kinetic energy: −f(4.0 m) = 0 − ½(2.0 kg)(6.0 m/s)² = −36 J. So f = 36 J / 4.0 m = 9.0 N.
11. A 1200 kg car moving at 20 m/s collides with an 800 kg car at rest, and the two cars lock together. What is their speed just after the collision?
- A 8.0 m/s
- B 10 m/s
- C 12 m/s
- D 15 m/s
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Answer: C. 12 m/s
Momentum is conserved in the collision: (1200)(20) = (1200 + 800)v, so v = 24,000 / 2000 = 12 m/s. Kinetic energy is not conserved when objects stick together, which is why setting the kinetic energies equal gives the wrong 15 m/s.
12. A 0.15 kg baseball moving at 40 m/s is hit straight back toward the pitcher at 50 m/s. The bat is in contact with the ball for 1.5 ms. What is the magnitude of the average force the bat exerts on the ball?
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Answer: 9.0 × 10³ N (9000 N)
Taking the ball's final direction as positive, its change in momentum is 0.15 × [50 − (−40)] = 13.5 kg·m/s. The average force is that impulse divided by the contact time: 13.5 / 0.0015 s = 9000 N.
13. A mechanic pulls on the end of a 0.25 m wrench with an 80 N force directed at 60° to the wrench handle. What is the magnitude of the torque about the bolt?
- A 17 N·m
- B 10 N·m
- C 20 N·m
- D 23 N·m
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Answer: A. 17 N·m
Torque is τ = rF sin θ, where θ is the angle between the handle and the force, so τ = 0.25 × 80 × sin 60° ≈ 17 N·m. Using cos 60° gives 10 N·m, and ignoring the angle gives 20 N·m, the torque for a perpendicular pull.
14. A solid sphere, a solid cylinder, and a thin hoop are released together from rest at the top of the same incline and roll without slipping. Which one reaches the bottom first?
- A The thin hoop
- B The solid cylinder
- C The solid sphere
- D All three arrive at the same time
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Answer: C. The solid sphere
For rolling without slipping, a = g sin θ / (1 + I/(MR²)), and I/(MR²) is 2/5 for the sphere, 1/2 for the cylinder, and 1 for the hoop. The sphere puts the smallest share of its energy into rotation, so it wins regardless of its mass or radius.
15. A 0.50 kg block on a frictionless horizontal surface is attached to a spring with k = 200 N/m. What is the period of its oscillation?
- A 0.31 s
- B 0.050 s
- C 0.16 s
- D 3.2 s
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Answer: A. 0.31 s
For a mass on a spring, T = 2π√(m/k) = 2π√(0.50/200) = 2π(0.050) ≈ 0.31 s. The period does not depend on the amplitude, and 3.2 is the frequency in hertz, not the period.