1. A block of mass 4 kg is moving with a velocity of 8 m/s. What is its
momentum?
A. 32 kg·m/s
B. 36 kg·m/s
C. 40 kg·m/s
D. 44 kg·m/s
Answer: A) 32 kg·m/s
Rationale: Momentum is given by p=mvp = mvp=mv, where m=4 kgm
= 4 \, \text{kg}m=4kg and v=8 m/sv = 8 \, \text{m/s}v=8m/s. So,
p=4×8=32 kg\cdotpm/sp = 4 \times 8 = 32 \,
\text{kg·m/s}p=4×8=32kg\cdotpm/s.
2. A particle is moving along a straight line with constant velocity.
What is its acceleration?
A. 0 m/s²
B. 1 m/s²
C. 9.8 m/s²
D. Undefined
Answer: A) 0 m/s²
Rationale: If the particle is moving with constant velocity, there is no
change in speed or direction, meaning the acceleration is zero.
,3. A car travels along a curved track with radius RRR. The car moves
at a constant speed vvv. What is the expression for the centripetal
acceleration of the car?
A. vR\frac{v}{R}Rv
B. v2R\frac{v^2}{R}Rv2
C. Rv\frac{R}{v}vR
D. v22R\frac{v^2}{2R}2Rv2
Answer: B) v2R\frac{v^2}{R}Rv2
Rationale: The centripetal acceleration is given by the formula a=v2Ra
= \frac{v^2}{R}a=Rv2, where vvv is the speed and RRR is the radius
of the curve.
4. A body is thrown horizontally from the top of a building with an
initial velocity of 8 m/s. How far will it travel horizontally in 4
seconds?
A. 32 m
B. 64 m
C. 40 m
D. 16 m
Answer: A) 32 m
Rationale: The horizontal distance is given by d=vtd = vtd=vt, where
v=8 m/sv = 8 \, \text{m/s}v=8m/s and t=4 st = 4 \, \text{s}t=4s.
So, d=8×4=32 md = 8 \times 4 = 32 \, \text{m}d=8×4=32m.
, 5. A car accelerates uniformly from rest to a speed of 25 m/s in 10
seconds. What is its acceleration?
A. 2.5 m/s²
B. 5 m/s²
C. 10 m/s²
D. 15 m/s²
Answer: A) 2.5 m/s²
Rationale: The acceleration is given by a=v−uta = \frac{v -
u}{t}a=tv−u, where v=25 m/sv = 25 \, \text{m/s}v=25m/s, u=0u =
0u=0, and t=10 st = 10 \, \text{s}t=10s. So,
a=25−010=2510=2.5 m/s2a = \frac{25 - 0}{10} = \frac{25}{10} =
2.5 \, \text{m/s}^2a=1025−0=1025=2.5m/s2.
6. A block of mass 4 kg slides down a frictionless incline of angle
30∘30^\circ30∘. What is the acceleration of the block down the incline?
A. 4.9 m/s24.9 \, \text{m/s}^24.9m/s2
B. 5.0 m/s25.0 \, \text{m/s}^25.0m/s2
C. 3.0 m/s23.0 \, \text{m/s}^23.0m/s2
D. 9.8 m/s29.8 \, \text{m/s}^29.8m/s2
Answer: B) 5.0 m/s25.0 \, \text{m/s}^25.0m/s2
Rationale: The acceleration down the incline is given by a=gsinθa =
g \sin\thetaa=gsinθ, where g=9.8 m/s2g = 9.8 \,
\text{m/s}^2g=9.8m/s2 and θ=30∘\theta = 30^\circθ=30∘.
momentum?
A. 32 kg·m/s
B. 36 kg·m/s
C. 40 kg·m/s
D. 44 kg·m/s
Answer: A) 32 kg·m/s
Rationale: Momentum is given by p=mvp = mvp=mv, where m=4 kgm
= 4 \, \text{kg}m=4kg and v=8 m/sv = 8 \, \text{m/s}v=8m/s. So,
p=4×8=32 kg\cdotpm/sp = 4 \times 8 = 32 \,
\text{kg·m/s}p=4×8=32kg\cdotpm/s.
2. A particle is moving along a straight line with constant velocity.
What is its acceleration?
A. 0 m/s²
B. 1 m/s²
C. 9.8 m/s²
D. Undefined
Answer: A) 0 m/s²
Rationale: If the particle is moving with constant velocity, there is no
change in speed or direction, meaning the acceleration is zero.
,3. A car travels along a curved track with radius RRR. The car moves
at a constant speed vvv. What is the expression for the centripetal
acceleration of the car?
A. vR\frac{v}{R}Rv
B. v2R\frac{v^2}{R}Rv2
C. Rv\frac{R}{v}vR
D. v22R\frac{v^2}{2R}2Rv2
Answer: B) v2R\frac{v^2}{R}Rv2
Rationale: The centripetal acceleration is given by the formula a=v2Ra
= \frac{v^2}{R}a=Rv2, where vvv is the speed and RRR is the radius
of the curve.
4. A body is thrown horizontally from the top of a building with an
initial velocity of 8 m/s. How far will it travel horizontally in 4
seconds?
A. 32 m
B. 64 m
C. 40 m
D. 16 m
Answer: A) 32 m
Rationale: The horizontal distance is given by d=vtd = vtd=vt, where
v=8 m/sv = 8 \, \text{m/s}v=8m/s and t=4 st = 4 \, \text{s}t=4s.
So, d=8×4=32 md = 8 \times 4 = 32 \, \text{m}d=8×4=32m.
, 5. A car accelerates uniformly from rest to a speed of 25 m/s in 10
seconds. What is its acceleration?
A. 2.5 m/s²
B. 5 m/s²
C. 10 m/s²
D. 15 m/s²
Answer: A) 2.5 m/s²
Rationale: The acceleration is given by a=v−uta = \frac{v -
u}{t}a=tv−u, where v=25 m/sv = 25 \, \text{m/s}v=25m/s, u=0u =
0u=0, and t=10 st = 10 \, \text{s}t=10s. So,
a=25−010=2510=2.5 m/s2a = \frac{25 - 0}{10} = \frac{25}{10} =
2.5 \, \text{m/s}^2a=1025−0=1025=2.5m/s2.
6. A block of mass 4 kg slides down a frictionless incline of angle
30∘30^\circ30∘. What is the acceleration of the block down the incline?
A. 4.9 m/s24.9 \, \text{m/s}^24.9m/s2
B. 5.0 m/s25.0 \, \text{m/s}^25.0m/s2
C. 3.0 m/s23.0 \, \text{m/s}^23.0m/s2
D. 9.8 m/s29.8 \, \text{m/s}^29.8m/s2
Answer: B) 5.0 m/s25.0 \, \text{m/s}^25.0m/s2
Rationale: The acceleration down the incline is given by a=gsinθa =
g \sin\thetaa=gsinθ, where g=9.8 m/s2g = 9.8 \,
\text{m/s}^2g=9.8m/s2 and θ=30∘\theta = 30^\circθ=30∘.
