JEE Advanced Physics 1st, 2nd & 3rd Laws of Motion PYQ

Q 1 :

A block of mass 5 kg moves along the $x$ -direction subject to the force $F = (- 20 x + 10)$ N, with the value of $x$ in metre. At time $t = 0$ s, it is at rest at position $x = 1 m.$ The position and momentum of the block at $t = (\frac{π}{4})$ s are [2024]

−0.5 m, 5 kg m/s
0.5 m, 0 kg m/s
0.5 m, −5 kg m/s
−1 m, 5 kg m/s

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(3)

Given mass of block = 5kg moving along the x-direction subject to the force $F = (- 20 x + 10)$ N with the value of $x$ in metre.

$Acceleration a = \frac{F}{m} \overset{F = (- 20 x + 10) N}{\leftarrow} m = 5 kg$

$t = 0, v = 0, x = 1 m$

$= \frac{- 20 x + 10}{5} = - 4 x + 2$

$Also, a = v \frac{d v}{d x} = - 4 x + 2$

$∴$ $\int_{0}^{v} v d v = \int_{1}^{x} (- 4 x + 2) d x$ $\Rightarrow \frac{v^{2}}{2} = {(- 2 x^{2} + 2 x)}_{1}^{x}$

$or, v = - 2 \sqrt{x - x^{2}} [since particle starts moving in -ve x-direction]$

$∴$ $\frac{d x}{d t} = - 2 \sqrt{x - x^{2}}$ $\Rightarrow \int_{x = 1}^{x = x} \frac{d x}{\sqrt{x - x^{2}}} = - 2 \int_{0}^{\frac{π}{4}} d t$

$\Rightarrow \sin^{- 1} {(2 x - 1)}_{1}^{x} = - \frac{π}{2}$

$∴$ $Position x = 0.5 m$

$And since v = - 2 \sqrt{x - x^{2}} = - 2 \sqrt{0.5 - {(0.5)}^{2}} = - 1 m/s$

$∴$ $Momentum P = m v = 5 (- 1) = - 5 {kg ms}^{- 1}$

Q 2 :

A particle of mass $m$ is moving in the $x y$ -plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = α (y \hat{x} + 2 x \hat{y}),$ where $α$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle? [2023]

$\vec{F} = 2 m α^{2} (x \hat{x} + y \hat{y})$
$\vec{F} = m α^{2} (y \hat{x} + 2 x \hat{y})$
$\vec{F} = 2 m α^{2} (y \hat{x} + x \hat{y})$
$\vec{F} = m α^{2} (x \hat{x} + 2 y \hat{y})$

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(1)

$∵$ $\vec{v} = α (y \hat{x} + 2 x \hat{y})$

$∴$ $\vec{a} = \frac{d \vec{v}}{d t} = α (\frac{d y}{d t} \hat{x} + 2 \frac{d x}{d t} \hat{y})$

$= α (v_{y} \hat{x} + 2 v_{x} \hat{y})$

$= α (2 x α \hat{x} + 2 α y \hat{y})$

$= 2 α^{2} [x \hat{x} + y \hat{y}]$

Q 3 :

A particle moves in the $X - Y$ plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ , where $A$ and $k$ are constants. The angle between the force and the momentum is [2007]

$0 °$
$30 °$
$45 °$
$90 °$

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(4)

$Given: momentum \vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$

$And, force \vec{F} = \frac{d \vec{p}}{d t} = A k [- \hat{i} \sin (k t) - \hat{j} \cos (k t)]$

$Here, \vec{F} \cdot \vec{p} = 0 But \vec{F} \cdot \vec{p} = F p \cos θ$

$∴$ $\cos θ = 0 \Rightarrow θ = 90 °$

$Hence, angle between the force and momentum, θ= 90 ° .$

Q 4 :

Statement-1: It is easier to pull a heavy object than to push it on a level ground and

Statement-2: The magnitude of frictional force depends on the nature of the two surfaces in contact. [2008]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Statement-1 is True, Statement-2 is False
Statement-1 is False, Statement-2 is True

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(2)

It is easier to pull a heavy object than to push it on a level ground. This is because the normal reaction in the case of pulling is less as compared by pushing. $(f = μ N)$ . Therefore the frictional force is small in case of pulling. The magnitude of frictional force depends on the nature of the two surfaces in contact. But is not the correct explanation of statement-1.

Q 5 :

Statement-1: A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table.

Statement-2: For every action there is an equal and opposite reaction. [2007]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Statement-1 is True, Statement-2 is False
Statement-1 is False, Statement-2 is True

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(2)

Cloth can be pulled out without dislodging the dishes from the table because of inertia.
Law of inertia is the Newton's first law of motion.
For every action there is an equal and opposite reaction.
This is Newton's third law of motion.

Topic Question Set

Q 1 :

A block of mass 5 kg moves along the $x$ -direction subject to the force $F = (- 20 x + 10)$ N, with the value of $x$ in metre. At time $t = 0$ s, it is at rest at position $x = 1 m.$ The position and momentum of the block at $t = (\frac{π}{4})$ s are [2024]

Q 2 :

A particle of mass $m$ is moving in the $x y$ -plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = α (y \hat{x} + 2 x \hat{y}),$ where $α$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle? [2023]

Q 3 :

A particle moves in the $X - Y$ plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ , where $A$ and $k$ are constants. The angle between the force and the momentum is [2007]

Q 4 :

Statement-1: It is easier to pull a heavy object than to push it on a level ground and

Statement-2: The magnitude of frictional force depends on the nature of the two surfaces in contact. [2008]

Q 5 :

Statement-1: A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table.

Statement-2: For every action there is an equal and opposite reaction. [2007]

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Topic Question Set

Q 1 : A block of mass 5 kg moves along the x-direction subject to the force F=(-20x+10)N, with the value of x in metre. At time t=0s, it is at rest at position x=1m. The position and momentum of the block at t=(π4) s are [2024]

Q 2 : A particle of mass m is moving in the xy-plane such that its velocity at a point (x,y) is given as v→=α(yx^+2xy^), where α is a non-zero constant. What is the force F→ acting on the particle? [2023]

Q 3 : A particle moves in the X-Y plane under the influence of a force such that its linear momentum is p→(t)=A[i^cos(kt)-j^sin(kt)], where A and k are constants. The angle between the force and the momentum is [2007]

Q 4 : Statement-1: It is easier to pull a heavy object than to push it on a level ground and Statement-2: The magnitude of frictional force depends on the nature of the two surfaces in contact. [2008]

Q 5 : Statement-1: A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table. Statement-2: For every action there is an equal and opposite reaction. [2007]

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Q 1 :

A block of mass 5 kg moves along the $x$ -direction subject to the force $F = (- 20 x + 10)$ N, with the value of $x$ in metre. At time $t = 0$ s, it is at rest at position $x = 1 m.$ The position and momentum of the block at $t = (\frac{π}{4})$ s are [2024]

Q 2 :

A particle of mass $m$ is moving in the $x y$ -plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = α (y \hat{x} + 2 x \hat{y}),$ where $α$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle? [2023]

Q 3 :

A particle moves in the $X - Y$ plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ , where $A$ and $k$ are constants. The angle between the force and the momentum is [2007]

Q 4 :

Statement-1: It is easier to pull a heavy object than to push it on a level ground and

Statement-2: The magnitude of frictional force depends on the nature of the two surfaces in contact. [2008]

Q 5 :

Statement-1: A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table.

Statement-2: For every action there is an equal and opposite reaction. [2007]