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Question 11 Mark

Consider a book lying on a table. The weight of the book and the normal force by the table on the book are equal in magnitude and opposite in direction. Is this an example of Newton's third law?

Answer

Yes, this is an example of Newton’s third law of motion, which sates that every action has an equal and opposite reaction.

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Question 21 Mark

An object is placed far away from all the objects that can exert force on it. A frame of reference is constructed by taking the origin and axes fixed in this object. Will the frame be necessarily inertial?

Answer

Yes, if the force on the object is zero, its acceleration w.r.t. all the other objects will we zero. So, the frame will necessarily be an inertial frame.

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Question 31 Mark

Raindrops of radius 1mm and mass 4mg are falling with a speed of 30m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head.

Answer

$\mathrm{r}=1 \mathrm{~mm}=10^{-3}$
$' \mathrm{~m}^{\prime}=4 \mathrm{mg}=4 \times 10^{-6} \mathrm{~kg}$
$\mathrm{~s}=10^{-3} \mathrm{~m} .$
$\mathrm{v}=0$
$\mathrm{u}=30 \mathrm{~m} / \mathrm{s}$
So, $\mathrm{a}=\frac{\mathrm{v}^2-\mathrm{u}^2}{2 \mathrm{~s}}=\frac{-30 \times 30}{2 \times 10^{-3}}=-4.5 \times 10^5 \mathrm{~m} / \mathrm{s}^2$ (decelerating)
Taking magnitude only deceleration is $4.5 \times 10^5 \mathrm{~m} / \mathrm{s}^2$
So, force $F=4 \times 10^{-6} \times 4.5 \times 10^5=1.8 \mathrm{~N}$

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Question 41 Mark

The acceleration of a particle is zero as measured from an inertial frame of reference. Can we conclude that no force acts on the particle?

Answer

No. The acceleration of the particle can also be zero if the vector sum of all the forces is zero, i.e. no net force acts on the particle.

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Question 51 Mark

Two blocks A and B of mass $m_A$ and $m_B$ respectively are kept in contact on a frictionless table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A?

Answer

Let, F → contact force between $m_A$ & $m_B$. And, f → force exerted by experimenter.

$\left.F+m_A a-f=0 m_B a-f=0 \Rightarrow F=f-m_A a \ldots \text {... } i\right) \Rightarrow F=m_B a \ldots \text {...(ii) From eqn. (i) and eqn. (ii) } \Rightarrow f-m_A a=m_B a \Rightarrow f=$
$m_B a+m_A a \Rightarrow f=a\left(m_A+m_B\right)$
$\Rightarrow\text{f}=\frac{\text{F}}{\text{m}_{\text{B}}}(\text{m}_{\text{B}}+\text{m}_{\text{A}})=\text{F}\Big(1+\frac{\text{m}_{\text{A}}}{\text{m}_{\text{B}}}\Big)$
$\Big[\text{because }\text{a}=\frac{\text{F}}{\text{m}_{\text{B}}}\Big]$
$\therefore$ The force exerted by the experimenter is $\text{F}\Big(1+\frac{\text{m}_{\text{A}}}{\text{m}_{\text{B}}}\Big)$

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