[2 Mark Question Answer]

Question 12 Marks

Why is gold not weighed by a spring balance?

Answer

Spring balance measures the weight of a body. Weight of a body changes from place to place on the surface of earth due to change in value of acceleration due to gravity. That is why gold is not weighed by a spring balance.

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Question 22 Marks

Derive the relation between newton and dyne.

Answer

Relation between newton and dyne :
$\begin{array}{l}1 N=1 kg x ms ^{-2}=1000 g \times 100 cm S ^{\sim}{ }^2 \\ 1 N=100000 g cm s ^{\prime 2} \\ 1 N=10^5 g cm s ^{-2} \\ 1 N=10^5 \quad\left[\therefore 1 \text { dyne }=1 g cm s ^{-2}\right]\end{array}$

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Question 32 Marks

State two factors which determine the momentum of a body.

Answer

Factors which determine the momentum of a body are:
1. Mass of the moving body: Larger the mass of a body, larger will be its
momentum.
2. Velocity of the body: Larger the velocity of the body, larger will be its momentum.

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Question 42 Marks

What do you understand by the term momentum?

Answer

Momentum: The instantaneous force which body possesses due to combined effect of mass and velocity is called momentum of the body. Mathematically, momentum is the product of mass and velocity of body.
P = mv
where P represents the momentum of the body.

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Question 52 Marks

What do you understand by the following terms?
(a) Contact forces
(b) Non-contact forces

Answer

(a) The forces which act on bodies when they are in actual contact, are called the contact forces. Two examples are as below:
1. Force of friction
2. Force of tension.
(b) The forces which act on bodies without being physically touched, are called the non- contact forces or the forces at a distance. The gravitational force, electrostatic force and magnetic force are the non-contact forces.

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Question 62 Marks

A body of mass 200 g is moving with a velocity of $5 ms^{-1}$. If the velocity of the body changes to $17 ms^{-1}$, calculate the change in linear momentum of the body.

Answer

Mass of body $= m =200 g=0.2 kg$
Velocity $= v _1=5 ms^{-} ;$Velocity $= v _2=17 ms^{-1}$
Change in linear momentum of body
$
=mv_2-m v_1=m\left(v_2-v_1\right)
$
$
\begin{array}{l}
=0.2(17-5) \\
=2.4 Ns or^{-1} kg ms^{-1}
\end{array}
$

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Question 72 Marks

An electron of mass $9 \times 10^{31} kg$ is moving with a linear velocity of $6 \times 10^7 ms^{-1}$. Calculate the linear momentum of electron.

Answer

$\begin{array}{l}\text { Mass of electron }= m =9 \times 10^{-31} kg \\ \text { Velocity of electron }= v =6 \times 10^7 ms^{-1} \\ \text { Momentum of electron }= P =\text { ? } \\ \text { We know, } P = mv \\ P =9.1 \times 10^{-31} \times 6 \times 10^7 \\ P =54 \times 10^{-24} kg ms ^{-1} \text { or } Ns \\ =54 \times 10^{-24} kg ms ^{-1} Ns \end{array}$

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Question 82 Marks

Calculate the velocity of a body of mass 0.5 kg , when it has a linear momentum of 5 Ns .

Answer

Mass of body $= m =0.5 kg$
Linear momentum $= P =5 Ns$
Velocity of body $= v =$ ?
We know $P = mv$
$
\begin{array}{r}
y =\frac{P}{ m }=\frac{5}{0.5} \\
v=10 ms^{-1}
\end{array}
$

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