PHY 111 FINAL EXAM PRACTICE EVALUATION 2026
2026 QUESTION BANK FULL SOLUTION SET
QUESTION 1:
A car is traveling at 10 m/s.
a. How fast would the car have to go to double its kinetic energy?
b. By what factor does the car's kinetic energy increase if its speed is doubled to 20 m/s?
Answer:
a. 14.14 m/s
b. 4 times
Rationale:
a. KE = ½mv². At 10 m/s, KE = ½m(100) = 50m. To double KE: 100m = ½mv² → v² = 200
→ v = 14.14 m/s
b. KE ∝ v², so doubling speed (×2) increases KE by factor of 2² = 4
QUESTION 2:
A 100 g ball is thrown straight up from a height of 1.5 m above the floor and just
reaches a ceiling 10 m above the floor. What was the initial speed of the ball?
Answer:
12.9 m/s
Rationale:
Using v² = v₀² + 2a(Δy), with v = 0 at maximum height:
0 = v₀² + 2(-9.8)(8.5)
,v₀² = 166.6
v₀ = 12.9 m/s
QUESTION 3:
A student throws a rock straight up with an initial speed of 7 m/s. What is the maximum
height relative to the student's hand? (Assume negligible air resistance)
Answer:
2.5 m
Rationale:
v² = v₀² + 2a(Δy)
0 = (7)² + 2(-9.8)(Δy)
Δy = 49/19.6 = 2.5 m
QUESTION 4:
A student at the top of a 20 m tall building drops a ball off the roof. How much time will
it take to hit the ground? What is the velocity just before impact?
Answer:
Time: 2.0 s; Velocity: -19.6 m/s
Rationale:
Using y = ½at² + v₀t + y₀:
0 = ½(-9.8)t² + 0 + 20 → t = 2.0 s
v = at + v₀ = (-9.8)(2.0) = -19.6 m/s
,QUESTION 5:
Compared to the initial speed of a rock thrown straight up, the final speed when it
returns to the student's hand will be:
a) larger b) smaller c) the same
Answer:
c) The same
Rationale:
With negligible air resistance, motion is symmetric. The distance up equals distance
down, and acceleration is uniform, so speeds are equal
QUESTION 6:
Calculate the power required for a 1,400 kg car to climb a 10° hill at a steady 80 km/hr.
Answer:
70.9 hp (5.289 × 10⁴ W)
Rationale:
80 km/hr = 22.2 m/s
P = F(v_avg) = (mg sin θ)(v)
P = (1400)(9.8)(sin 10°)(22.2) = 5.289 × 10⁴ W
P = 5.289 × 10⁴ / 746 = 70.9 hp
QUESTION 7:
Calculate the power required for a 1,400-kg car to accelerate from 90 to 110 km/hr in 6
seconds on a level road. Assume retarding force is 700 N.
, Answer:
82 hp (6.12 × 10⁴ W)
Rationale:
vᵢ = 90 km/hr = 25 m/s; v_f = 110 km/hr = 30.6 m/s
a_x = (30.6 - 25.0)/6.0 = 0.93 m/s²
F = ma_x + F_r = (1400)(0.93) + 700 = 2000 N
P = Fv = (2000)(30.6) = 6.12 × 10⁴ W = 82 hp
QUESTION 8:
What is the most accurate statement concerning work?
Answer:
Work on a system causes a change in the energy of the system
Rationale:
The work-energy theorem states that work done on a system equals the change in its
energy
QUESTION 9:
A 0.07 kg arrow is fired at 90 m/s at a 6.1 kg object initially at rest. The arrow emerges at
0.78 m/s. What is the resulting velocity of the object? Is the collision elastic or inelastic?
Answer:
a. 1.02 m/s; b. Inelastic
Rationale:
Using conservation of momentum:
2026 QUESTION BANK FULL SOLUTION SET
QUESTION 1:
A car is traveling at 10 m/s.
a. How fast would the car have to go to double its kinetic energy?
b. By what factor does the car's kinetic energy increase if its speed is doubled to 20 m/s?
Answer:
a. 14.14 m/s
b. 4 times
Rationale:
a. KE = ½mv². At 10 m/s, KE = ½m(100) = 50m. To double KE: 100m = ½mv² → v² = 200
→ v = 14.14 m/s
b. KE ∝ v², so doubling speed (×2) increases KE by factor of 2² = 4
QUESTION 2:
A 100 g ball is thrown straight up from a height of 1.5 m above the floor and just
reaches a ceiling 10 m above the floor. What was the initial speed of the ball?
Answer:
12.9 m/s
Rationale:
Using v² = v₀² + 2a(Δy), with v = 0 at maximum height:
0 = v₀² + 2(-9.8)(8.5)
,v₀² = 166.6
v₀ = 12.9 m/s
QUESTION 3:
A student throws a rock straight up with an initial speed of 7 m/s. What is the maximum
height relative to the student's hand? (Assume negligible air resistance)
Answer:
2.5 m
Rationale:
v² = v₀² + 2a(Δy)
0 = (7)² + 2(-9.8)(Δy)
Δy = 49/19.6 = 2.5 m
QUESTION 4:
A student at the top of a 20 m tall building drops a ball off the roof. How much time will
it take to hit the ground? What is the velocity just before impact?
Answer:
Time: 2.0 s; Velocity: -19.6 m/s
Rationale:
Using y = ½at² + v₀t + y₀:
0 = ½(-9.8)t² + 0 + 20 → t = 2.0 s
v = at + v₀ = (-9.8)(2.0) = -19.6 m/s
,QUESTION 5:
Compared to the initial speed of a rock thrown straight up, the final speed when it
returns to the student's hand will be:
a) larger b) smaller c) the same
Answer:
c) The same
Rationale:
With negligible air resistance, motion is symmetric. The distance up equals distance
down, and acceleration is uniform, so speeds are equal
QUESTION 6:
Calculate the power required for a 1,400 kg car to climb a 10° hill at a steady 80 km/hr.
Answer:
70.9 hp (5.289 × 10⁴ W)
Rationale:
80 km/hr = 22.2 m/s
P = F(v_avg) = (mg sin θ)(v)
P = (1400)(9.8)(sin 10°)(22.2) = 5.289 × 10⁴ W
P = 5.289 × 10⁴ / 746 = 70.9 hp
QUESTION 7:
Calculate the power required for a 1,400-kg car to accelerate from 90 to 110 km/hr in 6
seconds on a level road. Assume retarding force is 700 N.
, Answer:
82 hp (6.12 × 10⁴ W)
Rationale:
vᵢ = 90 km/hr = 25 m/s; v_f = 110 km/hr = 30.6 m/s
a_x = (30.6 - 25.0)/6.0 = 0.93 m/s²
F = ma_x + F_r = (1400)(0.93) + 700 = 2000 N
P = Fv = (2000)(30.6) = 6.12 × 10⁴ W = 82 hp
QUESTION 8:
What is the most accurate statement concerning work?
Answer:
Work on a system causes a change in the energy of the system
Rationale:
The work-energy theorem states that work done on a system equals the change in its
energy
QUESTION 9:
A 0.07 kg arrow is fired at 90 m/s at a 6.1 kg object initially at rest. The arrow emerges at
0.78 m/s. What is the resulting velocity of the object? Is the collision elastic or inelastic?
Answer:
a. 1.02 m/s; b. Inelastic
Rationale:
Using conservation of momentum:
