Kinematics - NEET Physics Chapterwise MCQs & PYQs

NEET Kinematics MCQs & PYQs

Practice NEET Kinematics Questions

Question 281:

easy

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R).

Assertion (A): If two particles, moving along straight line with constant velocities have to meet, the relative velocity must be along the line joining the two particles.

Reason (R): Relative motion means motion of one particle as viewed from the other particle.

 

For two particles to meet, the relative velocity vector must align with the line joining them so that from one's frame, the other moves directly towards it.

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Question 282:

moderate

Equation of trajectory of a projectile is \(y = \sqrt{3}x – 5x^2\). Then angle of projection with vertical is (Assume x-axis as horizontal and y-axis as vertical)

Comparing with \(y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}\), we find \(\tan\theta = \sqrt{3}\), so horizontal angle \(\theta = 60^\circ\). The angle with vertical is \(90^\circ - 60^\circ = 30^\circ\).

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Question 283:

easy

Two particles are projected with same initial velocity one makes angle \(\theta\) with horizontal while other makes an angle \(\theta\) with vertical. If their common range is R then product of their time of flight is directly proportional to:

(1999)

Short Solution

Concept: Time of flight for complementary angles and range formula.
Formula: \(T = (2u sin \alpha) / g\), \(R = (u^2 sin 2\alpha) / g\).
For angles \(\theta\) and \(90° - \theta\), times are \(T_1 = (2u sin \theta) / g\) and \(T_2 = (2u cos \theta) / g\).
Their product \(T_1 T_2 = (4u^2 sin \theta cos \theta) / g^2 = (2u^2 sin 2\theta) / g^2 = (2/g) R\). Thus, \(T_1 T_2 \propto R\).

Detailed Solution

Given:
Two particles are projected with the same initial velocity $u$.
Particle 1 is projected at an angle $\theta$ with the horizontal.
Particle 2 is projected at an angle $\theta$ with the vertical, meaning its angle of projection with the horizontal is $90^\circ - \theta$.
Both particles share the same horizontal range, $R$.
Formulas Used:
  • Time of flight: $T = \frac{2u \sin(\alpha)}{g}$
  • Horizontal Range: $R = \frac{u^2 \sin(2\alpha)}{g} = \frac{2u^2 \sin(\alpha)\cos(\alpha)}{g}$
Step-by-Step Derivation:
  1. Calculate the time of flight for the first particle ($T_1$):
    $$T_1 = \frac{2u \sin(\theta)}{g}$$
  2. Calculate the time of flight for the second particle ($T_2$):
    $$T_2 = \frac{2u \sin(90^\circ - \theta)}{g}$$
    Since $\sin(90^\circ - \theta) = \cos(\theta)$, this simplifies to:
    $$T_2 = \frac{2u \cos(\theta)}{g}$$
  3. Find the product of their times of flight ($T_1 T_2$):
    Multiply the two time equations together:
    $$T_1 T_2 = \left( \frac{2u \sin(\theta)}{g} \right) \left( \frac{2u \cos(\theta)}{g} \right)$$
    $$T_1 T_2 = \frac{4u^2 \sin(\theta)\cos(\theta)}{g^2}$$
  4. Relate the product to the horizontal range ($R$):
    Recall the expanded formula for the horizontal range $R$:
    $$R = \frac{2u^2 \sin(\theta)\cos(\theta)}{g}$$
    Rearrange the $T_1 T_2$ equation to group the components of the range formula:
    $$T_1 T_2 = \frac{2}{g} \left[ \frac{2u^2 \sin(\theta)\cos(\theta)}{g} \right]$$
    Substitute $R$ into the bracketed section:
    $$T_1 T_2 = \frac{2}{g} R$$
Because $\frac{2}{g}$ is a constant, the mathematical relationship shows that the product of the times of flight $T_1 T_2$ is directly proportional to the horizontal range $R$.
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Question 284:

difficult

A point P consider at contact point of a wheel on ground which rolls on ground without slipping then value of displacement of point P when wheel completes half of rotation (If radius of wheel is $1\text{ m}$):

(2002)

Short Solution:

Horizontal displacement is $\pi R$ and vertical displacement is $2R$.

Total displacement is given by $\sqrt{(pi R)^2 + (2R)^2} = R\sqrt{\pi^2+4}$, which evaluates to $\sqrt{pi^2+4}\text{ m}$ for $R = 1\text{ m}$.

Detailed Solution

Understanding the Physics (Rolling Without Slipping)

When a circular body like a wheel rolls without slipping, it simultaneously undergoes two types of motion:
  1. Translational Motion: The center of mass of the wheel moves forward in a straight line.
  2. Rotational Motion: The wheel spins around its central axis.
To find the net displacement of a specific point on the rim (point P), we must calculate its final horizontal and vertical positions relative to its starting point.

Step-by-Step Solution

1. Calculate the Horizontal Displacement (x-axis)
In one full rotation, a wheel covers a horizontal distance equal to its total circumference, which is $2\pi R$.
Because the wheel only completes half a rotation, the horizontal distance traveled by the center of the wheel is half of the circumference.
$$x = \pi R$$
2. Calculate the Vertical Displacement (y-axis)
Initially, point P is at the very bottom of the wheel, in direct contact with the ground. As the wheel completes exactly half a rotation (180 degrees), point P rotates from the bottom to the very top of the wheel.
Therefore, the vertical height gained by point P is equal to the wheel's diameter:
$$y = 2R$$
3. Determine the Net Displacement
Displacement is the shortest straight-line path between the initial and final positions. We can treat the horizontal and vertical displacements as perpendicular vectors and use the Pythagorean theorem to find the total displacement ($s$):
$$s = \sqrt{x^2 + y^2}$$
Substitute the values of $x$ and $y$ into the formula:
$$s = \sqrt{(\pi R)^2 + (2R)^2}$$
$$s = \sqrt{\pi^2 R^2 + 4R^2}$$
$$s = R\sqrt{\pi^2 + 4}$$
4. Substitute the Given Radius
The problem states that the radius of the wheel is $R = 1$ m. Substituting this value into our final equation gives:
$$s = 1 \cdot \sqrt{\pi^2 + 4}$$
$$s = \sqrt{\pi^2 + 4}$$

Final Answer

The total displacement of point P when the wheel completes half a rotation is $\sqrt{\pi^2 + 4}$ meters.

 

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