1. A ball is thrown horizontally with a speed of 12 m/s from a height of
5 m. How far will it travel horizontally before hitting the ground?
(Take g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
A. 20 m
B. 24 m
C. 30 m
D. 35 m
Answer: B) 24 m
Rationale: First, find the time to fall: t=2hg=2×510=1=1 st =
\sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = \sqrt{1} = 1
\, \text{s}t=g2h=102×5=1=1s.
Then, the horizontal distance is d=u×t=12×1=12 md = u \times t = 12
\times 1 = 12 \, \text{m}d=u×t=12×1=12m.
2. A stone is dropped from a height of 10 m. How long does it take to
reach the ground? (Take g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
A. 1 s
B. 2 s
C. 3 s
D. 4 s
Answer: B) 2 s
,Rationale: The time to fall is given by t=2hg=2×1010=2≈1.41 st =
\sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 10}{10}} = \sqrt{2}
\approx 1.41 \, \text{s}t=g2h=102×10=2≈1.41s, rounded down to 2
seconds.
3. A block of mass 5 kg is placed on a rough horizontal surface with a
coefficient of friction μ=0.3\mu = 0.3μ=0.3. What is the frictional force
acting on the block?
A. 15 N
B. 20 N
C. 12 N
D. 30 N
Answer: A) 15 N
Rationale: The frictional force is given by
Ffriction=μmgF_{\text{friction}} = \mu mgFfriction=μmg, where
μ=0.3\mu = 0.3μ=0.3, m=5 kgm = 5 \, \text{kg}m=5kg, and
g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2. So,
Ffriction=0.3×5×9.8=15 NF_{\text{friction}} = 0.3 \times 5 \times
9.8 = 15 \, \text{N}Ffriction=0.3×5×9.8=15N.
4. A block of mass 3 kg is moving in a circular path with a constant
speed of 6 m/s. The radius of the path is 4 m. What is the centripetal
force acting on the block?
A. 9 N
B. 18 N
C. 24 N
, D. 36 N
Answer: B) 18 N
Rationale: The centripetal force is given by F=mv2rF =
\frac{mv^2}{r}F=rmv2, where m=3 kgm = 3 \, \text{kg}m=3kg,
v=6 m/sv = 6 \, \text{m/s}v=6m/s, and r=4 mr = 4 \,
\text{m}r=4m. So, F=3×624=3×364=18 NF = \frac{3 \times
6^2}{4} = \frac{3 \times 36}{4} = 18 \,
\text{N}F=43×62=43×36=18N.
5. A body is moving in a circular path of radius 2 m with a constant
speed of 4 m/s. What is its centripetal acceleration?
A. 4 m/s²
B. 8 m/s²
C. 16 m/s²
D. 2 m/s²
Answer: B) 8 m/s²
Rationale: The centripetal acceleration is given by a=v2ra =
\frac{v^2}{r}a=rv2, where v=4 m/sv = 4 \, \text{m/s}v=4m/s and
r=2 mr = 2 \, \text{m}r=2m. So, a=422=162=8 m/s2a =
\frac{4^2}{2} = \frac{16}{2} = 8 \,
\text{m/s}^2a=242=216=8m/s2.
6. A body is moving in a circular path of radius rrr. The angular
velocity of the body is ω\omegaω. What is the relationship between
the linear velocity vvv and angular velocity ω\omegaω?
5 m. How far will it travel horizontally before hitting the ground?
(Take g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
A. 20 m
B. 24 m
C. 30 m
D. 35 m
Answer: B) 24 m
Rationale: First, find the time to fall: t=2hg=2×510=1=1 st =
\sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = \sqrt{1} = 1
\, \text{s}t=g2h=102×5=1=1s.
Then, the horizontal distance is d=u×t=12×1=12 md = u \times t = 12
\times 1 = 12 \, \text{m}d=u×t=12×1=12m.
2. A stone is dropped from a height of 10 m. How long does it take to
reach the ground? (Take g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
A. 1 s
B. 2 s
C. 3 s
D. 4 s
Answer: B) 2 s
,Rationale: The time to fall is given by t=2hg=2×1010=2≈1.41 st =
\sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 10}{10}} = \sqrt{2}
\approx 1.41 \, \text{s}t=g2h=102×10=2≈1.41s, rounded down to 2
seconds.
3. A block of mass 5 kg is placed on a rough horizontal surface with a
coefficient of friction μ=0.3\mu = 0.3μ=0.3. What is the frictional force
acting on the block?
A. 15 N
B. 20 N
C. 12 N
D. 30 N
Answer: A) 15 N
Rationale: The frictional force is given by
Ffriction=μmgF_{\text{friction}} = \mu mgFfriction=μmg, where
μ=0.3\mu = 0.3μ=0.3, m=5 kgm = 5 \, \text{kg}m=5kg, and
g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2. So,
Ffriction=0.3×5×9.8=15 NF_{\text{friction}} = 0.3 \times 5 \times
9.8 = 15 \, \text{N}Ffriction=0.3×5×9.8=15N.
4. A block of mass 3 kg is moving in a circular path with a constant
speed of 6 m/s. The radius of the path is 4 m. What is the centripetal
force acting on the block?
A. 9 N
B. 18 N
C. 24 N
, D. 36 N
Answer: B) 18 N
Rationale: The centripetal force is given by F=mv2rF =
\frac{mv^2}{r}F=rmv2, where m=3 kgm = 3 \, \text{kg}m=3kg,
v=6 m/sv = 6 \, \text{m/s}v=6m/s, and r=4 mr = 4 \,
\text{m}r=4m. So, F=3×624=3×364=18 NF = \frac{3 \times
6^2}{4} = \frac{3 \times 36}{4} = 18 \,
\text{N}F=43×62=43×36=18N.
5. A body is moving in a circular path of radius 2 m with a constant
speed of 4 m/s. What is its centripetal acceleration?
A. 4 m/s²
B. 8 m/s²
C. 16 m/s²
D. 2 m/s²
Answer: B) 8 m/s²
Rationale: The centripetal acceleration is given by a=v2ra =
\frac{v^2}{r}a=rv2, where v=4 m/sv = 4 \, \text{m/s}v=4m/s and
r=2 mr = 2 \, \text{m}r=2m. So, a=422=162=8 m/s2a =
\frac{4^2}{2} = \frac{16}{2} = 8 \,
\text{m/s}^2a=242=216=8m/s2.
6. A body is moving in a circular path of radius rrr. The angular
velocity of the body is ω\omegaω. What is the relationship between
the linear velocity vvv and angular velocity ω\omegaω?
