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Phy 111 Final Exam Practice Evaluation 2026 Question Bank Full Solution Set

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PHY 111 FINAL EXAM PRACTICE EVALUATION 2026 QUESTION BANK FULL SOLUTION SET

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PHY 111 FINAL EXAM PRACTICE
EVALUATION 2026 QUESTION BANK
FULL SOLUTION SET
PHY 111 FINAL EXAM PRACTICE EVALUATION 2026
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Instructions: Choose the single best answer for each question.
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1. Kinematics (1D Motion)
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A car accelerates from rest at a constant rate of 2.0 m/s22.0m/s2 for 5.0 seconds. What is the t
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otal distance traveled by the car during this time?
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• A) 10 m hg hg




• B) 25 m hg hg




• C) 50 m hg hg




• D) 100 m hg hg




• Correct Answer: B (Using x=v0t+12at2=0+12(2)(5)2=25 mx=v0t+21at2=0+21
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(2)(5)2=25m)



2. Kinematics (Projectile Motion)
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A ball is thrown horizontally from the top of a 45-meter-
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tall cliff with a speed of 10 m/s10m/s. Ignoring air resistance, approximately how long does it t
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ake for the ball to hit the ground? (g=9.8 m/s2g=9.8m/s2)
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• A) 2.1 s hg hg




• B) 3.0 s hg hg




• C) 4.6 s hg hg




• D) 9.0 s hg hg




• Correct Answer: B (Time depends only on vertical motion: 45=12(9.8)t2⇒t≈3.03 s45=2
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1(9.8)t2⇒t≈3.03s)

,3. Newton's Laws (Forces)
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A 5.0 kg box is pulled across a frictionless horizontal surface by a horizontal force of 20 N. Wha
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t is the acceleration of the box?
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• A) 0.25 m/s20.25m/s2
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• B) 4.0 m/s24.0m/s2
hg




• C) 10 m/s210m/s2
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• D) 100 m/s2100m/s2
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• Correct Answer: B (Using F=ma⇒a=F/m=20/5=4.0 m/s2F=ma⇒a=F/m=20/5=4.0m/s2)
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4. Newton's Laws (Friction)
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A 10 kg crate is at rest on a horizontal floor. The coefficient of static friction is 0.40 and the coe
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fficient of kinetic friction is 0.30. What horizontal force is required to start the crate moving?
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• A) 29.4 N hg hg




• B) 39.2 N hg hg




• C) 98.0 N hg hg




• D) 392 N hg hg




• Correct Answer: B (Maximum static friction: fsmax=μsFN=μsmg=0.40×10×9.8=39.2 Nfs
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max=μsFN=μsmg=0.40×10×9.8=39.2N)



5. Work and Energy
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A force of 50 N is used to push a box 4.0 meters across a floor at a constant velocity. How muc
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h work is done by the force?
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• A) 12.5 J hg hg




• B) 50 J hg hg




• C) 200 J hg hg




• D) 0 J hg hg




• Correct Answer: C (W=Fdcos⁡θ=50×4×cos⁡(0∘)=200 JW=Fdcosθ=50×4×cos(0∘)=200J
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)

,6. Work-Energy Theorem
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A 2.0 kg object is moving at 3.0 m/s3.0m/s. What is its kinetic energy?
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• A) 3.0 J
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• B) 6.0 J
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• C) 9.0 J
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• D) 18 J
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• Correct Answer: C (KE=12mv2=12(2)(3)2=9 JKE=21mv2=21(2)(3)2=9J)
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7. Conservation of Energy
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A 1.0 kg ball is dropped from a height of 10 meters. Ignoring air resistance, what is its speed ju
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st before it hits the ground? (g=9.8 m/s2g=9.8m/s2)
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• A) 7.0 m/s
hg hg




• B) 9.8 m/s
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• C) 14 m/s
hg hg




• D) 98 m/s
hg hg




• Correct Answer: C (mgh=12mv2⇒v=2gh=2×9.8×10≈14 m/smgh=21mv2⇒v=2gh
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=2×9.8×10≈14m/s)



8. Momentum and Impulse
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A 0.5 kg ball is moving at 10 m/s10m/s and is brought to rest in 0.2 seconds by a force. What i
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s the magnitude of the average force exerted on the ball?
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• A) 1 N
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• B) 5 N
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• C) 25 N
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• D) 50 N
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• Correct Answer: C (Impulse = Change in momentum: FΔt=mΔv⇒F=0.5×10/0.2=25 NFΔ
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t=mΔv⇒F=0.5×10/0.2=25N)

, 9. Conservation of Momentum
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A 4.0 kg object moving at 6.0 m/s6.0m/s collides elastically with a stationary 2.0 kg object. If th
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e collision is perfectly elastic, which of the following is true?
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• A) Kinetic energy is conserved.
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• B) Momentum is not conserved.
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• C) The objects stick together after the collision.
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• D) The 4.0 kg object stops completely.
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• Correct Answer: A (By definition, kinetic energy is conserved in an elastic collision.)
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10. Rotational Kinematics
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A wheel rotates with a constant angular acceleration of 2.0 rad/s22.0rad/s2. If it starts from rest,
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what is its angular velocity after 3.0 seconds?
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• A) 1.5 rad/s
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• B) 3.0 rad/s
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• C) 6.0 rad/s
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• D) 9.0 rad/s
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• Correct Answer: C (ω=ω0+αt=0+(2)(3)=6.0 rad/sω=ω0+αt=0+(2)(3)=6.0rad/s)
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11. Torque hg




A wrench is used to tighten a bolt. A force of 20 N is applied perpendicular to the wrench at a
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distance of 0.25 m from the bolt. What is the torque applied?
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• A) 0.25 N·m
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• B) 5.0 N·m
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• C) 20 N·m
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• D) 80 N·m
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• Correct Answer: B (τ=rFsin⁡θ=0.25×20×sin⁡(90∘)=5.0 N⋅mτ=rFsinθ=0.25×20×sin(90∘)
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=5.0N⋅m)

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