Physics M.C.Q (1 Marks)

2. $\frac{\text{u}}{\cos\theta}$

Explanation:

Along the string, the velocity of each object is the same.

$2\text{v}\cos(\theta)=2\text{u}$

$\text{v}=\frac{\text{u}}{\cos(\theta)}$

1. $\text{v}=\frac{\text{v}_1+\text{v}_2}{2}$

Explanation:

Velocity is uniform in both cases; that is, acceleration is zero.

We have:

$\text{d}_1=\text{v}_1\text{t}$ and $\text{d}_2=\text{v}_2\text{t}$

Total displacement, $\text{d = d}_1+\text{d}_2$

Total time, $\text{t = t + t = 2t}$

$\therefore$ Average velocity, $\text{v}=\frac{\text{d}_1+\text{d}_2}{2\text{t}}=\frac{\text{v}_1+\text{v}_2}{2}$

3. $\frac{2}{\text{v}}=\frac{1}{\text{v}_1}+\frac{1}{\text{v}_2}$

Explanation:

Velocity is uniform in both cases; that is, acceleration is zero.

$\text{x = v}_1\text{t}_1\Rightarrow\text{t}_1=\frac{\text{x}}{\text{v}_1}$

$\text{x = v}_2\text{t}_2\Rightarrow\text{t}_2=\frac{\text{x}}{\text{v}_2}$

Total displacement, $\text{x}'=\text{2x}$

Total time, $\text{t}=\text{t}_1+\text{t}_2$

$\therefore$ Average velocity, $\text{v}=\frac{\text{x}'}{\text{t}}=\frac{2\text{v}_1\text{v}_2}{\text{v}_1+\text{v}_2}$

$\Rightarrow\frac{2}{\text{v}}=\frac{1}{\text{v}_1}+\frac{1}{\text{v}_2}$

1. The particle has a constant acceleration.

4. The average speed in the interval 0 to 10s is the same as the average speed in the interval 10s to 20s.

Explanation:

1. The slope of the v-t graph gives the acceleration. For the given graph, the slope is constant. So, acceleration is constant.
2. From 0 to 10 seconds, velocity is in positive direction and then in negative direction. This means that the particle turns around at t = 10s.
3. Area in the v-tcurve gives the distance travelled by the particle.

Distance travelled in positive Direction ≠ Distance travelled in negative direction

$\therefore\text{Displacement}\neq\text{Zero}$

4. The area of the v-tgraph from t = 0s to t = 10s is the same as that from t = 10s to t = 20s. So, the distance covered is the same. Hence, the average speed is the same.

1. If the velocity and acceleration have opposite sign, the object is slowing down.

2. If the position and velocity have opposite sign, the particle is moving towards the origin.

4. If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

Explanation:

1. Acceleration is given by

$-\text{a}=\frac{\text{dv}}{\text{dt}}$

$-\text{a}<0$

$\Rightarrow\frac{\text{dv}}{\text{dt}}<0$

$\Rightarrow\text{V}_{\text{final}}<\text{V}_{\text{initial}}$

2. If the position and velocity have opposite signs, the particle moves towards the origin. It can be explained by following figure:

Object is moving toward the origin.

Object is moving toward the origin.

3. If the velocity is zero in a certain time interval, then the change in the velocity in that time interval will also be zero. As acceleration is rate of change of velocity, it will also be zero at an instant in that time interval.

4. The acceleration of S₂ with respect to S₁ may be anything between zero and 8m/s.

Explanation:

$\overrightarrow{\text{a}}_{\text{s}_2\text{s}_1}=\overrightarrow{\text{a}}_{\text{s}_2\text{P}}+\overrightarrow{\text{a}}_{\text{Ps}_1}$

$\big|\overrightarrow{\text{a}}_{\text{s}_2\text{s}_1}|=\sqrt{4^2+4^2+32\cos(\theta)}$

$-1<\cos(\theta)<1$

$0<\overrightarrow{\text{a}}_{\text{s}_2\text{s}_1}<8$

1. The particle has come to rest 6 times.

Explanation:

1. The slope of the x-tgraph gives the velocity. Here, 6 times the slope is zero. So, the particle has come to rest 6 times.
2. As the slope is not maximum at t = 6s, the maximum speed is not at t = 6s.
3. As the slope is not positive from t = 0s to t = 6s, the velocity does not remain positive.

4. Average velocity $\frac{\text{Total displacement}}{\text{Total time taken}}=\frac{\text{x final}-\text{x initial}}{\text{t}}$

For the shown time (r = 6s), the displacement of the particle is positive. Therefore, the average velocity is positive.

3. The acceleration of the particle is 2a.
4. At t = 2s particle is at the origin.

Explanation:

Initial velocity $=\Big|\frac{\text{dx}}{\text{dt}}\Big|_{\text{t} = 0}$

$\frac{\text{dx}}{\text{dt}}=\text{u}+2\text{a}(\text{t}-2\text{s})$

$\Big|\frac{\text{dx}}{\text{dt}}\Big|_{\text{t}=0}=\text{u}-4\text{as}\neq\text{u}$

Acceleration $=\frac{\text{d}^2\text{x}}{\text{dt}^2}=2\text{a}$

At t = 2s,

x = u(2s - 2s) + a(2s - 2s)² = 0 (origin)

1. The magnitude of the velocity of a particle is equal to its speed.

Explanation:

1. Velocity being a vector quantity has magnitude as well as direction, and magnitude of velocity is called speed.
2. Average velocity $=\frac{\text{Total displacement}}{\text{Total time taken}}$

Average speed $=\frac{\text{Total distance travelled}}{\text{Total time taken}}$

$\text{Distance}\geq\text{Displacement}$

$\therefore\text{Average speed}\geq\text{Average velocity}$

The magnitude of average velocity in an interval is not always equal to its average speed in that interval.

3. If speed is always zero, then the distance travelled is always zero. Hence, the total distance travelled and the average speed will be zero.
4. If the speed of a particle is never zero, the distance travelled by the particle is never zero. Hence, the average speed will not be zero.

4. 100m.

Explanation:

For the same urange, $\text{R}\propto\sin(2\theta).$

So,

$\frac{\text{R}_1}{\text{R}_2}=\frac{\sin(2\theta_1)}{\sin(2\theta_2)}$

$\Rightarrow\text{R}_2=50\times\frac{\sin(90)}{\sin(30)}=100\text{m}$

1. The displacement is zero.

4. The average velocity is zero.

Explanation:

Displacement is zero because the initial and final positions are the same.

$\text{Average velocity}=\frac{\text{Displacement}}{\text{Total time}} =0$

$\text{Distance covered}=2\pi\text{r}\neq0$

$\text{Average speed}=\frac{\text{Distance travelled}}{\text{Total time taken}}\neq0$

4. The information is insufficient to decide the relation of R_A with R_B.

Explanation:

Horizontal range for the projectile, $\text{R}=\frac{\text{u}^2\sin(2\text{a})}{\text{g}}$

Information of the initial velocity is not given in the question.

3. v_A = v_B

Explanation:

Total energy of any particle $=\frac{1}{2}\text{mv}^2+\text{mgh}$

Both the particles were at the same height and thrown with equal initial velocities, so their initial total energies are equal. By the law of conservation of energy, their final energies are equal.

At the ground, they are at the same height. So, their P.E. are also equal; this implies that their K.E. should also be equal. In other words, their final velocities are equal.

1. Average speed of a particle in a given time is never less than the magnitude of the average velocity.
2. It is possible to have a situation in which $\Big|\frac{\text{d}\vec{\text{v}}}{\text{dt}}\Big|\neq0$ but $\frac{\text{d}}{\text{dt}}|\vec{\text{v}}|=0.$
3. The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.

Explanation:

1. Average speed of a particle in a given time is never less than the magnitude of the average velocity.
2. It is possible to have a situation in which $\Big|\frac{\text{d}\vec{\text{v}}}{\text{dt}}\Big|\neq0$ but $\frac{\text{d}}{\text{dt}}|\vec{\text{v}}|=0.$
3. The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.

Example, the motion of a particle on a circular track with a constant speed.

Average velocity $=\frac{\text{Displacement}}{\text{Total time}}$

$\text{Displacement}\leq\text{Distance}$

$\therefore\text{Average velocrty}\leq\text{Average speed}$

In uniform circular motion, speed is constant but velocity is not.

$\text{i.e.},\Big|\frac{\text{d}\vec{\text{v}}}{\text{dt}}\Big|\neq0$ but $\frac{\text{d}}{\text{dt}}=|\vec{\text{v}}|=0$ which proves case (b)

4. In one complete circle of uniform motion, average velocity is zero. Instantaneous velocity is never zero in the interval.

3. Both will reach simultaneously.

Explanation:

Because the downward acceleration and the initial velocity in downward direction of the two bullets are the same, they will take the same time to hit the ground and for a half projectile.

Time of flight $=\text{T}=\sqrt{\frac{2\text{h}}{\text{g}}}$

2. 70km/h towards south-west

Explanation:

Final velocity, $\overrightarrow{\text{V}}_{\text{f}}=-50\hat{\text{i}}\text{km/h}$

Initial velocity, $\overrightarrow{\text{V}}_{\text{i}}=50\hat{\text{j}}\text{km/h}$

Change in velocity, $\triangle\overrightarrow{\text{V}}=\overrightarrow{\text{V}}_{\text{f}}-\overrightarrow{\text{V}}_{\text{i}}$

$|\triangle\text{V}|=\sqrt{50^2+50^2+2\times50\times50\cos(90^{\circ})}=70\text{km/h}$

It is towards southwest, as shown in the figure.

4. The information is insufficient to decide the relation of x_A with x_B.

Explanation:

As velocity and acceleration are in opposite directions, velocity will become zero after some time (t) and the particle will return.

$\therefore0=\text{u}-\text{at}$

$\Rightarrow\text{t}=\frac{\text{u}}{\text{a}}$

Because the value of acceleration is not given, we cannot say that the particle will return after/ before 10 seconds.