Q # 1. What is the difference between uniform and variable velocity? From the explanation of variable velocity, define acceleration. Give the SI unit of velocity and acceleration.
Uniform Velocity: A body is said to have a uniform velocity if it covers equal displacement in equal intervals of time.
Variable Velocity: A body is said to have a variable velocity if it covers unequal displacements in equal intervals of time.
Acceleration: The time rate of change of velocity of the body is called acceleration. Consider a body is moving with initial velocity $\mathbf{v}_{\mathbf{i}}$ and after some time $\Delta t$ its velocity becomes $\mathbf{v}_{\mathbf{f}}$, then the acceleration a of the object will be:
$$\mathbf{a}=\frac{\mathbf{v}_{\mathbf{f}}-\mathbf{v}_{\mathbf{i}}}{t}$$
SI Unit of Velocity: The SI unit of velocity is meter per second or $\mathrm{ms}^{-1}$
SI Unit of Acceleration: The SI unit of velocity is meter per second per second or $\mathrm{ms}^{-2}$
Q # 2. An object is thrown vertically upward. Discuss the sign of acceleration due to gravity, relative to velocity, while the object is) in air.
Ans. When the object is thrown vertically upward, it will move against the direction of gravity. The sign of acceleration $\mathbf{g}$ relative to velocity $\mathbf{v}$ will be taken as negative. It is because of the reason that the direction of $\mathbf{g}$ is opposite to the direction of $\mathbf{v}$ during upward motion.
If the object is moving downward, then the sign of $\mathbf{g}$ relative to $\mathbf{v}$ will be taken as positive because both $\mathbf{g}$ and $\mathbf{v}$ are in same direction.
Q # 3. Can the velocity of an object reverse the direction when the acceleration is constant? If so, give an example.
Ans. Yes, the velocity of a body can reverse its direction with constant acceleration. For example, when a body is thrown vertically upward under the action of gravity, the velocity of the object will go on decreasing because force of gravity is acting downward.
When the object reaches the maximum height, its velocity becomes zero, and then the object reverses its direction of motion and start moving vertically downward. During the whole process, the magnitude of the acceleration due to gravity remains constant.
Q # 4. Specify the correct statement:
a. An object can have a constant velocity even its speed is changing.
b. An object can have a constant speed even its velocity is changing.
c. An object can have a zero velocity even its acceleration is not zero.
d. An object subjected to a constant acceleration can reverse its velocity.
Ans. The statement (b) is correct.
An object can have constant speed even its velocity is changing. For the case of circular motion, the object moves with constant speed but its velocity changes due to change in direction continuously.
Q # 5. A man standing on the top of a tower throws a ball straight up with initial velocity $v_{i}$ and at the same time throws a second ball straight downward with the same speed. Which ball will have a larger speed when it strikes the ground? Ignore the air friction. Ans. Both balls will hit the ground with same speed.
Ans. When a ball is thrown upward with initial velocity $\mathbf{y}_{\mathbf{i}}$, if willhave same velocity $\mathbf{v}_{\mathbf{i}}$ when it returns back to the same level. After that the ball will continue its motion in downward direction and hits the ground with velocity $\mathbf{v}_{\mathbf{f}}$.
Thus if the second ball is thrown vertically downward with initial velocity $\mathbf{v}_{\mathbf{i}}$ from the same height, it will hit the ground with the same finalvelocity $\mathbf{v}_{\mathbf{f}}$.
Q # 6. Explain the circumstances in which the velocity $v$ and acceleration $\mathbf{a}$ of a car are
(i) Parallel
(ii) Anti-parallel
(iii) Perpendicular to one another
(iv) $v$ is zero but a is not zero
(v) $\mathbf{a}$ is zero but $v$ is not zero
Ans.
(i) When the velocity of the car is increasing along a straight line then $\mathbf{v}$ and $\mathbf{a}$ of the car will be parallel to each other.
(ii) When the velqcity of the car is decreasing along a straight line then $\mathbf{v}$ and $\mathbf{a}$ of the car will be anti-parallel to each other.
(iii) Whep the car moves along circular path, then a will be directed towards the center of the circle while its velocity will be along the tangent. Thus $\mathbf{v}$ and $\mathbf{a}$ of the car will be perpendicular to each other when it moves on a circular path.
(iv) When the brake is applied on a moving car, it slows down and comes to rest due to negative acceleration in opposite direction. Thus $\mathbf{v}$ is zero but $\mathbf{a}$ is not zero.
(v) When the car is moving in straight line with uniform velocity, then $\mathbf{a}$ of the car is zero but $\mathbf{v}$ is not zero.
Q # 7. Motion with constant velocity is a special case of motion with constant acceleration. Is this statement is true? Discuss.
Ans. Yes this statement is true. When a body moves with constant velocity in the straight line, its acceleration is zero. Hence, the acceleration of the body will always remains constant during such motion. As the zero is a constant quantity, therefore this is a special case of motion.
Q # 8. Find the change in momentum for an object subjected to a given force for a given time and state the law of motion in terms of momentum.
Ans. Consider a body of mass $m$ is moving with an initial velocity $\mathbf{v}_{i}$. Suppose an external force $\mathbf{F}$ acts upon it for time $t$ after which the velocity becomes $\mathbf{v}_{f}$. The acceleration $\mathbf{a}$ prodyced by this force is given by:
$$\mathbf{a}=\frac{\mathbf{v}_{f}-\mathbf{v}_{i}}{t}$$
By Newton's second law, acceleration is given as
$$\mathbf{a}=\frac{\mathbf{F}}{m}$$
Equating both equations, we get
$$ \frac{\mathbf{F}}{m}=\frac{\mathbf{v}_{f}-\mathbf{v}_{i}}{t}$$
$$ \Rightarrow \mathbf{F}=\frac{m\left(\mathbf{v}_{f}-\mathbf{v}_{i}\right)}{t}=\frac{m \mathbf{v}_{f}-m \mathbf{v}_{i}}{t}$$
$$\Rightarrow \mathbf{F} =\frac{\mathbf{p}_{f}-\mathbf{p}_{i}}{t}$$
where $\mathbf{p}_{i}=m \mathbf{v}_{i}$ and $\mathbf{p}_{f}=m \mathbf{v}_{f}$ are the initial and final momentum of the body.
$$\Rightarrow \mathbf{F}=\frac{\Delta \mathbf{p}}{t}$$
$\because \mathbf{p}_{f}-\mathbf{p}_{i}=\Delta \mathbf{p}=$ Change in linear momentum
This is Newton's second law of motion in terms of linear momentum.
Statement: The time rate of change of momentum of a body is equal to the applied force.
Q # 9. Define impulse and show that how it is related to linear momentum?
Ans.
Impulse: When a force is acted on a body for a very short time $\Delta t$, the product of force and time is called impulse It is a vector quantity and its unit is $\mathrm{N}$ s. Mathematically it is described as:
$$\mathbf{I}=\mathbf{F} \times \Delta t$$
Where I is the impulse of force $\mathbf{F}$.
Relationship between Impulse and Momentum
According to the Newton's second law of motion, the rate of change of linear momentum is equal to the applied force. Mathematically it is described as:
$$\begin{aligned} & \mathbf{F}=\frac{\Delta \mathbf{p}}{\Delta t} \\ & \Rightarrow \mathbf{F} \times \Delta t=\Delta \mathbf{p} \end{aligned} $$
As Impulse $\mathbf{I}=\mathbf{F} \times \Delta t$
Therefore, the equation (1) will become:
$$\mathbf{I}=\Delta \mathbf{p}$$
Hence Impulse = Change in Linear Momentum
Q # 10. State the law of conservation of linear momentum, pointing out the importance of an isolated system. Explain, why under certain conditions, the law is useful even though the system is not completely isolated?
Ans.
Statement: The total linear momentum of an isolated system remains constant.
Isolated System: It is a system on which no external agency exerts any force. In an isotated system, the bodies may interact with each other but no external force acts on them. Thus, in an isolated system, the linear momentum of the system remains conserve.
Explanation: In ever day life, the effect of frictional forces and gravitational force is negligible. Thus law of conservation of momentum can be applied to the systems which are not completely isolated e.g., firing of gun, motion of rocket etc.
Q # 11. Explain the difference between elastic and inelastic collision. Explain how would a bouncing ball behave in each case? Give the plausible reason for the fact that K.E is conserved in most cases?
Ans.
Elastic collision: A collision in which the K.E. of the system is conserved is called elastic collision
Inelastic collision: A collision in which the K.E. of the system is not conserved is called inelastic collision.
Explanation: When a ball is dropped on floor, after the impact it attains the same height. It is because of the fact that small amount of K.E is converted into heat and sound energies.
Q # 12. Explain what is meant by projectile motion. Derive the expression for (a) Time of flight (b) Range of projectile. Show that the range of the projectile is maximum when the projectile is thrown at an angle of $45^{\circ}$ with the horizontal.
Projectile Motion: It is the two dimensional motion in which the object moves under constant acceleration due to gravity. During projectile motion, the object has constant horizontal component of velocity but changing vertical component of velocity.
Time of Flight: The time taken by the object to cover the distance from the place of its projection to the place where it hits the ground at the same level is called time of flight.
As the projectile goes up and comes back to the same level, thus covering no vertical distance i.e., $S=h=0$. Thus the time of flight $t$ can be find out by using $2^{\text {nd }}$ equation of motion:
$$\begin{aligned} & S=v_{i y} \times t+\frac{1}{2} a_{y} t^{2} \\ & \Rightarrow 0=v_{i} \sin \theta \cdot t-\frac{1}{2} g t^{2} \\ & \Rightarrow \frac{1}{2} g t^{2}=v_{i} \sin \theta \cdot t \\ & \Rightarrow t=\frac{2 v_{i} \sin \theta}{g} \end{aligned}$$
This is the expression of time of flight of a projectile.
Range of the Projectile: The distance which the projectile covers in the horizontal direction is called the range of the projectile. In projectile motion, the horizontal component of velocity remains same. Therefore the range $R$ of the projectile can be determine using formula:
$$R=v_{i x} \times t$$
where $v_{i x}$ is the horizontal component of velocity and $t$ is the time of flight of projectile.
$$\begin{aligned}& R=v_{i} \cos \theta \times\left(\frac{2 v_{i} \sin \theta} {g}\right)=\frac{v_{i}^{2}}{g} 2 \sin \theta \cos \theta \\ & \Longrightarrow R=\frac{v_{i}^{2}}{g} \sin 2 \theta \end{aligned}$$
Thus the range of projectile depends upon the velocity of projection and angle of projection.
Maximum Horizontal Range: The horizontal range will be maximum when the factor $\sin 2 \theta$ will be maximum. So,
Maximum value of $\sin 2 \theta=1$
$$\Rightarrow 2 \theta=\sin { }^{-1}(1)=90^{\circ}$$
$$\Rightarrow \theta=45^{\circ}$$
Hence for the maximum horizontal range, the angle of projection should be $45^{\circ}$.
Q # 13. At what point or points in its path does a projectile have its minimum speed, its maximum speed?
Ans. The speed of the projectile is minimum at the maximum height of projectile. It is because of the reason that, at maximum height the vertical component of velocity becomes zero.
The speed of the projectile is maximum at the point of projection and also just before it strikes the ground because the vertical component of velocity is maximum at these points.
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