105 practice questions covering Laws of Motion, organised into 0 topics.
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Sample questions
Q1
hard
A block of mass 2 kg is placed at the top of a rough incline making an angle of 60° with the horizontal. The coefficient of static friction is 0.5. Will the block remain at rest on the incline? (Take g = 9.8 m/s²)
AYes, because static friction is sufficient to prevent sliding.
BNo, because the static friction is too high.
CYes, because the angle of incline is steep enough for equilibrium.
DNo, because the component of gravitational force exceeds static friction.
Show answer and explanation
Correct answer: D - No, because the component of gravitational force exceeds static friction.
The force component down the incline is mgsin(θ) = 2*9.8*sin(60°)=16.97 N. Maximum static friction is μ_s*N = 0.5*2*9.8*cos(60°) = 4.9 N. Since 16.97 N > 4.9 N, the block will slide. The incorrect options reflect misunderstandings or calculation errors.
Q2
medium
A lamp of weight $60\ \text{N}$ hangs from the ceiling by a string. A horizontal force pulls the lamp aside until the string makes $30^\circ$ with the vertical, and the lamp rests there. What is the tension in the string?
A$69.3\ \text{N}$
B$120\ \text{N}$
C$52.0\ \text{N}$
D$60\ \text{N}$
Show answer and explanation
Correct answer: A - $69.3\ \text{N}$
Resolving vertically, $T\cos 30^\circ = W$, so $T = 60/0.866 = 69.3\ \text{N}$. The string now supports a sideways pull as well, so its tension must exceed the weight. $60\ \text{N}$ ignores the tilt, $52.0\ \text{N}$ multiplies by $\cos 30^\circ$ instead of dividing, and $120\ \text{N}$ resolves with $\sin 30^\circ$, using the wrong component.
Q3
hard
An astronaut floating in space throws a 2 kg wrench at 5 m/s. If the astronaut has a mass of 80 kg, what is their recoil velocity?
A0.125 m/s
B2 m/s
C-2 m/s
D-0.125 m/s
Show answer and explanation
Correct answer: D - -0.125 m/s
By conserving momentum: 0 = (2 kg)(5 m/s) + (80 kg)(v_astronaut). Solving gives v_astronaut = -0.125 m/s. Errors arise from sign confusion or incorrect momentum assignments.
Q4
medium
A crate of mass $20\ \text{kg}$ slides at constant velocity along a horizontal floor when pulled by a horizontal force of $50\ \text{N}$. What is the coefficient of kinetic friction between the crate and the floor? (Take $g = 10\ \text{m/s}^2$)
A$4.0$
B$0.025$
C$0.25$
D$2.5$
Show answer and explanation
Correct answer: C - $0.25$
At constant velocity the pull balances kinetic friction, so $f = 50\ \text{N}$, while $N = mg = 200\ \text{N}$. Hence $\mu_k = f/N = 50/200 = 0.25$. The value $2.5$ divides by the mass instead of the weight, $4.0$ inverts the ratio as $N/f$, and $0.025$ is a decimal slip. The coefficient is a pure number and carries no unit.
Q5
medium
When a passenger in a bus standing still suddenly feels a backward jerk as the bus starts moving, this is an example of:
ALaw of universal gravitation.
BNewton's third law of motion.
CNewton's second law of motion.
DNewton's first law of motion.
Show answer and explanation
Correct answer: D - Newton's first law of motion.
The backward jerk is due to inertia, as the passenger's body resists the change in motion when the bus accelerates. This illustrates Newton's first law. The other laws do not explain this observation.
Q6
hard
Two ice skaters, one with mass 60 kg and another with mass 40 kg, push off each other and move apart on a frictionless surface. If the 60 kg skater moves with a velocity of 2 m/s, what is the velocity of the 40 kg skater?
A2.5 m/s
B3 m/s
C4 m/s
D1.5 m/s
Show answer and explanation
Correct answer: B - 3 m/s
By conservation of momentum, the total initial momentum is zero. Let v be the velocity of the 40 kg skater. 0 = (60 kg)(2 m/s) + (40 kg)(v). Solving gives v = -3 m/s, meaning the 40 kg skater moves with 3 m/s in the opposite direction. Students might mistakenly use incorrect mass ratios or sign reversals.
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