Physics M.C.Q (1 Marks)

3. Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.

Explanation:

As shown in diagram AB and AC are two smooth planes inclined to the angle $\theta_1$ and $\theta_2$ respectively. As friction is absent here, hence, mechanical energy will be conserved. As both the tracks having common height h, From conservation of mechanical energy,

$\frac{1}{2}\text{mv}^2=\text{mgh}$ (for both tracks I and II)

$\text{v}=\sqrt{2\text{gh}}$

Hence, speed is same for both stones. For stone I, a₁ = acceleration along inclines plane $=\text{g}\sin\theta_1$

Similarly, for stone II, $\text{a}_2=\text{g}\sin\theta_2$ as $\theta_2>\theta_1$ hence, $\text{a}_2>\text{a}_1$

And both length for track II is also less, hence stone II reaches earkier than stone I.

3. Total mechanical energy.

Explanation:

As the body is falling freely under gravity, the potential energy decreases continuously and kinetic energy increases continuously as all the conservative forces are doing work. So, total mechanical energy (PE + KE) of the body will be constant.

Let us discuss this in detail:

In the given diagram an object is dropped from-a height H from ground.

At point A total mechanical energy will be EA = K.E + P.E

$\text{E}_\text{A}=\frac{1}{2}\text{mv}^2+\text{mgH}$

As velocity will be zero at A, so its kinetic energy will be zero.

$\text{E}_\text{A}=\text{mgH}$

Velocity at point B will be, $\text{v}_\text{B}=\sqrt{2\text{gh}}$

So, energy at point B will be $\text{E}_\text{B}=\text{KE}+\text{PE}$

$\text{E}_\text{B}=\frac{1}{2}\text{m}(2\text{gh})+\text{mg}(\text{H}-\text{h})$

$\text{E}_\text{B}=\text{mgh}+\text{mgH}-\text{mgh}$

$\text{E}_\text{B}=\text{mgH}$

Now, velocity at point C will be $\text{v}_\text{c}=\sqrt{2\text{gh}}$

So, energy at point will be $\text{E}_\text{c}=\text{KE}+\text{PE}$

$\text{E}_\text{c}=\frac{1}{2}\text{m}(2\text{gH})+\text{mg}(0)$

$\text{E}_\text{c}=\text{mgH}$

So, total mechanical energy will remain same (if we neglect the air friction).

2. The magnetic forces do no work on each particle.

Explanation:

Key concept: To calculate the change in kinetic energy of the system during motion we have to apply work energy, theorem. According to this theorem, Net work done = Final kinetic energy - Initial kinetic energy of the object The above statement shows the connection between work and kinetic energy as:

“The work done by the net force acting on an object is equal to the change in the kinetic energy of that object”.

Net work done (IF) on a particle equals change in kinetic energy of the particle.

$\sum\text{W}=\text{K}_2-\text{K}_1$

According to the problem as the electron and proton are moving under the influence of mutual forces, the magnetic forces will be perpendicular to their motion, hence, it acts as a centripetal force for the particle. In this way the particle performs the uniform circular morion, this implies speed will remain constant. So, there is no change in kinetic energy of the particle. Hence no work is done by these forces.

$\vec{\text{F}_\text{m}}=\text{q}(\vec{\text{v}}\times\vec{\text{B}})\cdot\text{F}_\text{m}$

(magnetic force) will be perpendicular to both B and v, where B is the external magnetic field and v is the velocity of particle. That is why one ignores the magnetic force of one particle on another.

1. $250\pi^2$

Explanation:

According to the problem, Radius = 1m, mass = m = 5kg

$\text{f}=\frac{300}{60}$

Angular velocity will be

$=2\pi\text{f}=(300\times2\pi)\text{rad/ min}$

$=(300\times3.14)\text{rad/ 60s}$

$=\frac{300\times2\times3.14}{60}\text{rad/ s}=10\pi\text{rad/ s}$

And relation between linear velocity and angular velocity is $\text{v}=\omega\text{R}$

$=\Big(\frac{300\times2\pi}{60}\Big)(1\text{m})$

$=10\pi\text{m/ s}$

And kinetic energy $=\frac{1}{2}\text{mv}^2$

$=\frac{1}{2}\times5\times(10\pi^2)$

$=100\pi^2\times5\times\frac{1}{2}$

$=250\pi^2\text{J}$

2. V = E, K = 0

Explanation:

Key concept: Energy Graph for a Spring: If the mass attached with spring performs simple harmonic motion about its mean position then its potential energy at any position (x) can be given by,

Now kinetic energy at any position $\text{K}=\text{E}-\text{U}=\frac{1}{2}\text{Ka}^2-\frac{1}{2}\text{Kx}^2$

$\text{K}=\frac{1}{2}(\text{a}^2-\text{x}^2)\ ....(\text{i})$

From the above formula, we can cheak that

$\text{U}_\text{max}=\frac{1}{2}\text{Ka}^2$ $\big[\text{At extreme x}=\pm\text{a}\big]$ and U_min = 0 [At mean x = 0]

$\text{K}_\text{max}=\frac{1}{2}\text{Ka}^2$ [At mean x = 0] and K_min = 0 $\big[\text{At extreme x}=\pm\text{a}\big]$

$\text{E}=\frac{1}{2}​​\text{ka}^2=​​\text{constant(at all positions)}$

Because velocity of mass = 0 [at extreme position]

$\therefore\ \text{K}=\frac{1}{2}\text{mv}^2=0$

It means kinetic energy changes parabolically w.r.t. position but total energy remain always constant irrespective to the position of mass.

Total energy is E = PE + KE, which remains constant.

According to the question, when paraticle is at x = xm, i.e., at extreme position x = xm ⇒ KE = 0

Hence, total energy will be

$\text{E}=\text{PE}+0=\text{PE}$

$\Rightarrow\text{V}\text{(x}_\text{m})=\frac{1}{2}\text{kx}^2_\text{m}$

Hence option (b) is correct.

V = E, K = 0

2. The velocity of the bullet will be more than half of its earlier velocity.

4. The bullet will move in a different parabolic path.

6. The internal energy of the particles of the target will increase.

Explanation:

1. Let KE₂, KE₁ are the kinetic energy of bullet before and after hitting and targrt,

$1\text{KE}_2=\frac{1}{2}\text{KE}_1$

$\frac{1}{2}\text{mv}^2_2=\frac{1}{2}\cdot\frac{1}{2}\text{mv}^2_1$

$\text{v}^2_2=\frac{1}{2}\text{v}^2_1=\Big(\frac{\text{v}_1}{\sqrt{2}}\Big)^2$

$=\Big(\frac{\text{v}_1\sqrt{2}}{2}\Big)^2=(0.707\text{v}_1)^2$

2. v₂ = 0.707v₁. Hence, the velocity of bullet after target is not reduce to half. If rejects option (a).
3. v₂ = 0.707v₁. So velocity of bullet after target is more than half of its earlier velocity verifies option (b).
4. Bullet has horizontal velocity so its path will be parabolic but with new parabola as both components vx and vy changes after emerging out from target. So rejects the option (c).
5. As above discussed path of bullet after target will be of new parabola, verifies the option (d).
6. As bullet has horizontal and vertical components so has new parabola of range smaller than previous. So rejects the option (e).
7. As some parts of kinetic energy of bullet converted into heat so internal energy o target increased. Verifies option (f).

4. 155.0J

Explanation:

If air resistance is negligible, total mechanical energy of the system will remain constant. And let us take ground as a reference where potential energy will be zero.

According to the problem, h = 1.5 m, v = 1m/ s, m = 10 kg, g = 10m s^-2

Initial energy of the shotput $=(\text{PE})_\text{i}+(\text{KE})_\text{i}$

$=\text{mgh}+\frac{1}{2}\text{mv}^2$

$=10\times10\times1.5+\frac{1}{2}\times10\times(1)^2$

$=150+5=155.0\text{J}$

From conservation of mechanical energy,

$\text{(PE)}_\text{i}+\text{(KE)}_\text{i}=\text{(PE)}_\text{f}+\text{(KE)}_\text{f}$

So, final kinetic energy of the shotput is 155J

2. 50J

Explanation:

Key concept: When a variable force acts on a particle while it moves from point A to B, say along the path shown in the figure, work done by the force on the particle is given by

On the particle is given by

$\text{W}=\int\limits^{\text{B}}_\text{A}\vec{\text{F}}\cdot\vec{\text{ds}}\ ...(\text{i})$

Hence, $\int\limits^{\text{B}}_\text{A}\vec{\text{F}}\cdot\vec{\text{ds}}$ is to be integrated along the path the particle follows.

The vector integral is equivalent to

$\text{W}=\int\limits^{\text{x}_2}_{\text{x}_1}\text{f}_\text{x}\text{dx}+\int\limits^{\text{y}_2}_{\text{y}_1}\text{f}_\text{y}\text{dy}+\int\limits^{\text{z}_2}_{\text{z}_1}\text{f}_\text{z}\text{dz}$

According to the problem, velocity = ax^3/2, mass = 0.5kg, a = 5m^-1/2s^-1

We have to find work done (W) by net force.

We know that,

Acceleration, $\text{a}_0=\frac{\text{dv}}{\text{dt}}=\text{v}\frac{\text{dv}}{\text{dx}}$

$=\text{ax}^{\frac{3}{2}}\frac{\text{d}}{\text{dx}}\Big(\text{ax}^{\frac{3}{2}}\Big)$

$=\text{ax}^{\frac{3}{2}}\times\text{a}\times\frac{3}{2}\times\text{x}^{\frac{1}{2}}=\frac{3}{2}\text{a}^2\text{x}^2$

Net Force $=\text{ma}_0=\text{m}\Big(\frac{3}{2}\text{a}^2\text{x}^2\Big)$

And work done due to variable force,

Work done $=\int\limits^{\text{x}=2}_{\text{x}=0}\text{F dx}=\int\limits^2_0\frac{3}{2}\text{ma}^2\text{x}^2\text{ dx}$

$=\frac{3}{2}\text{ma}^2\times\Big(\frac{\text{x}^3}{3}\Big)^2_0$

$=\frac{1}{2}\text{ma}^2\times8$

$=\frac{1}{2}\times(0.5)\times(25)\times8=50\text{J}$

Explanation:

For constant power

Displacement $\propto\text{t}^{\frac{3}{2}}$

Because, $\text{P}=\frac{\vec{\text{F}}\cdot\vec{\text{ds}}}{\text{dt}}=\vec{\text{F}}\cdot\vec{\text{v}}=\text{constant}$

$(\because$ P = constant according to the problem$)$

Now, will by dimensional analysis

[F][v] = constant

⇒ [MLT^-2][LT^-1] = constant

⇒ L²T^-3 = constant $(\because$ mass is constant$)$

$\Rightarrow\text{L}\propto\text{T}^{\frac{3}{2}}$

$\Rightarrow\text{ Displacement (d)}\propto\text{t}^{\frac{3}{2}}$

Explanation:

Key concept: In a collision if the motion of colliding particles before and after the collision is along the same line, the collision is said to be head on or one dimensional.

When two bodies of equal masses collides elastically, their velocities are interchanged. Kinetic energy and linear momentum remains conserved Total kinetic energy of the system before collision

$=\frac{1}{2}\text{mv}^2+0=\frac{1}{2}\text{mv}^2$

In (a), kinetic energy of the system after collision,

$\text{K}_1=\frac{1}{2}(2\text{m})\Big(\frac{\text{v}}{2}\Big)^2=\frac{1}{4}\text{mv}^2$

Hence this option is incorrect.

In (b), kinetic energy of the system after collision,

$\text{K}_2=\frac{1}{2}(\text{m})(\text{v})^2=\frac{1}{2}\text{mv}^2$

Hence this option will be correct.

In (c), kinetic energy of the system after collision,

$\text{K}_3=\frac{1}{2}(3\text{m})\Big(\frac{\text{v}}{3}\Big)^2=\frac{1}{6}\text{mv}^2$

Hence this option is incorrect.

In (d), kinetic energy of the system after collision,

$\text{K}_4=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{m}\Big(\frac{\text{v}}{2}\Big)^2+\frac{1}{2}\text{m}\Big(\frac{\text{v}}{3}\Big)^2=\frac{49}{72}\text{mv}^2$

We see that kinetic energy is conserves only in (b).

3. $1.05\times10^4\text{N}$

Explanation:

We know that force $\text{F}=\frac{\Delta\text{p}}{\Delta\text{t}}$ and $\frac{\Delta\text{p}}{\Delta\text{t}}=\frac{\text{m}[(-\text{v})-\text{u}]}{\text{t}}=\frac{-2\text{mv}}{\text{t}}$

And the magnitude of force will be, $\text{F}=\frac{2\text{mv}}{\text{t}}$

According to the problem, $\text{m}=150\text{g}=\frac{150}{1000}\text{kg}=\frac{3}{20}\text{kg}$

$\Delta\text{t}=\text{time of contact}=0.001\text{s}$

$\text{u}=126\text{km/ h}=\frac{126\times1000}{60\times60}\text{m/ s}=35\text{}\text{m s}$

$\text{v}=-126\text{km/ h}=-35\text{m/ s}$

Change in momentum of the ball

$\Delta\text{P}=\text{m(v}-\text{u})$

$=\frac{3}{20}(-70)=-\frac{21}{2}$

We know that force $\text{F}=\frac{\Delta\text{P}}{\Delta\text{t}}$

$=\frac{\frac{-21}{2}}{0.001}\text{N}=-1.05\times10^4\text{N}$

Here, -ve sign shows that force will be opposite to the direction of movement of the ball before hitting. So the force that the batsman had to apply to hold the bat firmly at its place would be F = 1.05 × 10⁴N K