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AP Physics 1 Unit 1A: Kinematics: 1D Motion, Position & Velocity — practice questions — page 2

Explore 87 free Unit 1A: Kinematics: 1D Motion, Position & Velocity questions in AP Physics 1 Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 1A: Kinematics: 1D Motion, Position & Velocity questions and exercises with answers

Showing questions 31–60 of 87.

The elastic potential energy stored in a spring of spring constant $k$ when stretched by displacement $x$ is:Read the answer and worked solution →The work-energy theorem states that the net work done on a body equals:Read the answer and worked solution →$\vec{P} + \vec{Q} = \vec{P} - \vec{Q}$ হলে [AIEEE 2002]Read the answer and worked solution →দুটি সমান মানের বল F পরস্পর সমকোণে ক্রিয়াশীল। মোট বলের মান [সংসদ নমুনা প্রশ্ন]Read the answer and worked solution →দুটি ভেক্টরের প্রতিটির মান P। এদের লম্বির মানও P। ভেক্টর দুটির মধ্যে পারস্পরিক কোণ হল [সংসদ নমুনা প্রশ্ন]Read the answer and worked solution →P এবং $\sqrt{2}P$ ভেক্টর দুটির লম্বি বৃহত্তরটির সঙ্গে Θ কোণে আনত। দেখাও যে, Θ $\neq$Read the answer and worked solution →$\vec{F} = (6\hat{i} - 8\hat{j} + 10\hat{k})$ N বল কোনো একটি ভরের ওপর ক্রিয়া করে 1 m s⁻¹ ত্বরণ সৃষ্টি করল। বস্তুটির ভর হল [CBSE PMT 1996]Read the answer and worked solution →একটি বিন্দুতে ক্রিয়ারত দুটি বলের যোগফল 16 N। যদি তাদের লম্বির অভিমুখ কম বলটির ওপর লম্ব এবং মান 8 N হয়, তাহলে বল দুটি হবে [Pb.PMT 1995]Read the answer and worked solution →একটি বল P অপর একটি বল P-এর সঙ্গে 20° কোণ উৎপন্ন করে। বলছয়ের লম্বি হবেRead the answer and worked solution →যদি $\vec{A} = \vec{B} + \vec{C}$ এবং $\vec{A}$, $\vec{B}$ ও $\vec{C}$-এর স্কেলার মান যথাক্রমে 5, 4 এবং 3 হয় তাহলে $\vec{A}$ এবং $\vec{C}$-এর মধ্যবর্তী কোণ হল [JEE'12]Read the answer and worked solution →$\vec{P}$ এবং $\vec{Q}$ যথাক্রমে পূর্বাভিমুখী এবং দক্ষিণ-পশ্চিমাভিমুখী। $\vec{R} = \vec{P} + \vec{Q}$ হলেRead the answer and worked solution →Two point particles of equal mass are placed at x = 2 m and x = 8 m on the x-axis. Where does the centre of mass of the system lie?Read the answer and worked solution →If $\hat{i}$, $\hat{j}$, $\hat{k}$ are right-handed orthogonal unit vectors, then $\hat{i} \times \hat{j}$ equals:Read the answer and worked solution →For a homogeneous (uniformly dense) thin circular disc, the centre of mass lies at:Read the answer and worked solution →Two friends push a heavy box from the same side, applying forces of 30 N and 40 N in the same direction. What is the net force on the box?Read the answer and worked solution →An **ideal fluid** has all the following properties EXCEPT:Read the answer and worked solution →The pressure $p$ at depth $h$ below the free surface of a liquid of density $\rho$ (in addition to atmospheric pressure $p_0$) is:Read the answer and worked solution →The **SI unit** of pressure is:Read the answer and worked solution →A **mercury barometer** measures:Read the answer and worked solution →**Reynolds number** is used to distinguish:Read the answer and worked solution →**Viscosity** of most liquids with rising temperature:Read the answer and worked solution →**Stokes' law** gives the viscous drag on a small sphere of radius $r$ moving with velocity $v$ through a fluid of viscosity $\eta$ as:Read the answer and worked solution →The phenomenon of rise (or fall) of a liquid in a narrow capillary tube is called:Read the answer and worked solution →The condition for **terminal velocity** of a sphere falling in a viscous fluid is that:Read the answer and worked solution →A swimmer dives to 6 m below the water surface. The gauge pressure on her (water density = 1000 kg/m³, $g = 9.8$ m/s²) is:Read the answer and worked solution →A hydraulic lift uses a small piston (area 1 cm²) and a large piston (area 100 cm²). To lift a 1000 kg load, the minimum force needed on the small piston is:Read the answer and worked solution →Two liquids of densities $\rho_1$ and $\rho_2$ exert the same pressure. The ratio of their column heights $h_1 : h_2$ equals:Read the answer and worked solution →A capillary tube of radius $r$ dipped in a liquid of surface tension $T$ and density $\rho$, contact angle $\theta$, gives rise (or fall) to height $h$ where:Read the answer and worked solution →A steel ball of radius 0.3 mm falls at 2 m/s through glycerine ($\eta = 0.833$ N·s/m²). The viscous drag is:Read the answer and worked solution →A liquid flowing through a tube has Reynolds number $R_n = 500$. Its flow is:Read the answer and worked solution →

Worked practice questions

The elastic potential energy stored in a spring of spring constant $k$ when stretched by displacement $x$ is:

  1. $kx$
  2. $\dfrac{1}{2} k x^2$
  3. $k x^2$
  4. $\dfrac{1}{2} k^2 x$
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Correct answer: B. $\dfrac{1}{2} k x^2$

Force needed to stretch a spring by $x$ is $F = kx$ (Hooke's law). Work done in stretching from 0 to $x$: $U = \int_0^x kx' \, dx' = \dfrac{1}{2} k x^2$. The $\tfrac{1}{2}$ shows up because force grows linearly during the stretch — average force is $kx/2$, total work is $kx/2 \cdot x = \tfrac{1}{2} k x^2$. Option A ($kx$) is the *force*, not the energy.

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The work-energy theorem states that the net work done on a body equals:

  1. The change in its kinetic energy
  2. The change in its potential energy
  3. The total mechanical energy
  4. The momentum of the body
Show answer and solution

Correct answer: A. The change in its kinetic energy

Work-energy theorem: $W_{\text{net}} = \Delta KE = KE_{\text{final}} - KE_{\text{initial}}$. This is a direct consequence of Newton's second law. Apply $F = ma$ and integrate over displacement: $\int F \, dx = \int m a \, dx = \int m v \, dv = \tfrac{1}{2} m v^2 \big|_{u}^{v} = \Delta KE$. If only conservative forces act, this is equivalent to saying $\Delta KE + \Delta PE = 0$, the conservation of mechanical energy. With friction or other dissipative forces present, energy is lost to heat and the strict mechanical-energy conservation breaks, but the work-energy theorem still holds because work includes *all* forces.

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$\vec{P} + \vec{Q} = \vec{P} - \vec{Q}$ হলে [AIEEE 2002]

  1. $\vec{P}$ ও $\vec{Q}$ পরস্পর সমান্তরাল
  2. $\vec{P}$ ও $\vec{Q}$ পরস্পর লম্ব
  3. $\vec{P} = 0$
  4. $\vec{Q} = 0$
Show answer and solution

Correct answer: D. $\vec{Q} = 0$

সমীকরণটি থেকে দুই পাশে $\vec{P}$ বাদ দিলে পাই $$\vec{Q}=-\vec{Q}$$ এখন দুই পাশে $\vec{Q}$ যোগ করলে পাই $$2\vec{Q}=0$$ অতএব $$\vec{Q}=0$$ এখন অপশনগুলো মিলিয়ে দেখি, এটি $(D)$-এর সঙ্গে মিলে।

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More Physics topics

Unit 8: Fluids: Density, Buoyancy and ContinuityUnit 1B: Kinematics: 2D Motion & ProjectilesUnit 1C: Kinematics: Relative & Circular MotionUnit 2A: Newton's Laws and Translational EquilibriumUnit 2B: Friction, Gravitational and Normal ForcesUnit 3A: Work and Kinetic Energy TheoremUnit 3B: Potential Energy & Conservation of EnergyUnit 4A: Linear Momentum and Impulse