Assertion (A): In any curved motion magnitude of dot product of unit acceleration vector & unit velocity vector \(|\hat{a} \cdot \hat{v}|\) cannot be equal to 1.
Reason (R): In all accelerated straight line motion \(|\hat{a} \cdot \hat{v}|\) cannot be less than 1.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
The magnitude of the dot product \(|\hat{a} \cdot \hat{v}| = |\cos\theta|\), where \(theta\) is the angle between \(vec{a}\) and \(vec{v}\). For curved motion, \(vec{a}\) and \(vec{v}\) are never parallel or anti-parallel (\(theta \ne 0^\circ, 180^\circ\)), so \(|\cos\theta| \ne 1\). Thus (A) is true. For straight line motion, \(vec{a}\) and \(vec{v}\) are always parallel or anti-parallel, so \(|\cos\theta| = 1\). Thus (R) is true and implies it cannot be less than 1, and (R) explains (A).
Assertion (A): Two stones are simultaneously projected from level ground from same point with same speeds but different angles with horizontal. Both stones move in same vertical plane. Then the two stones may collide in mid air.
Reason (R): For two stones projected simultaneously from same point with same speed at different angles with horizontal, their trajectories must intersect at some point except projection point.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
For collision, the projectiles must be at the same position at the same time. If launched simultaneously from the same point, their x-positions at time \(t\) are \(x_1 = u \cos\theta_1 t\) and \(x_2 = u \cos\theta_2 t\). For \(x_1 = x_2\) at \(t > 0\), \(cos\theta_1 = \cos\theta_2\), which means \(theta_1 = \theta_2\), contradicting 'different angles'. Hence, they cannot collide. So (A) is false. Trajectories \(y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}\) do intersect for \(0 < \theta_1, \theta_2 < 90^\circ\), but if extreme angles (\(0^\circ\) or \(90^\circ\)) are included, trajectories may not intersect beyond the origin. Thus, (R) is false in a general sense.
Assertion (A): The maximum range along the inclined plane, when thrown downward is greater than that when thrown upward along the same inclined plane with same speed at same angle from incline.
Reason (R): The maximum range along inclined plane is independent of angle of inclination.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
The maximum range down an incline is \(R_{\text{max, down}} = \frac{u^2}{g(1-\sin\alpha)}\) and up an incline is \(R_{\text{max, up}} = \frac{u^2}{g(1+\sin\alpha)}\). Since \(1-\sin\alpha 0\), \(R_{\text{max, down}} > R_{\text{max, up}}\). So (A) is true. Both formulas clearly depend on the angle of inclination \(\alpha\). Thus (R) is false.
Assertion (A): In projectile motion, speed always decreases.
Reason (R): In presence of air drag, projectile motion is a uniformly accelerated motion.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): In ideal projectile motion, speed decreases on the way up and increases on the way down, reaching a minimum at the peak. It does not always decrease. So (A) is False.
Reason (R): In the presence of air drag, the drag force depends on velocity, making the net acceleration non-constant. Thus, it is not uniformly accelerated motion. So (R) is False.
Since both (A) and (R) are false, option (4) is correct.
Assertion (A): When speed of projection of a body is made (n) times, its time of flight becomes (n) times.
Reason (R): At this speed, the range of projectile becomes (n^2) times.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): Time of flight \(T = \frac{2u sin\theta}{g}\). If (u) is replaced by (nu), (T' = nT). So (A) is True.
Reason (R): Horizontal range \(R = \frac{u^2 sin(2\theta)}{g}\). If (u) is replaced by (nu), (R' = n^2 R). So (R) is True.
Both statements are true. However, the scaling of range (R) does not explain the scaling of time of flight (A). They are independent consequences of initial speed scaling. So (R) is not the correct explanation for (A). Option (2) is correct.
Assertion (A): When the range of a projectile is maximum, the time of flight is the largest.
Reason (R): Horizontal range is maximum when angle of projection is (90^circ).
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): Maximum range occurs at (theta = 45^circ). The largest time of flight occurs at (theta = 90^circ). Since these angles are different, the assertion is false. So (A) is False.
Reason (R): Horizontal range is maximum at (theta = 45^circ) (when (sin(2theta)=1)). At (theta = 90^circ), the range is zero. So (R) is False.
Since both (A) and (R) are false, option (4) is correct.
Assertion (A): A particle is projected from ground on a horizontal plane with speed \(10\text{ ms}^{-1}\) and angle of projection \(37^\circ\) with horizontal. Its velocity vector will be perpendicular to initial velocity vector after \(\frac{4}{3}\text{ s}\).
Reason (R): Two vectors \(\vec{v}\) and \(\vec{u}\) are perpendicular then \(\vec{u} \cdot \vec{v} = 0\).
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): Initial velocity \(\vec{u} = (10 cos 37^\circ)\hat{i} + (10 sin 37^\circ)\hat{j} approx 8\hat{i} + 6\hat{j}\). Velocity at time (t) is \(\vec{v}(t) = 8\hat{i} + (6 - gt)\hat{j}\). For perpendicularity, \(\vec{u} \cdot \vec{v} = 0 ⇒ 64 + 6(6 - gt) = 0 ⇒ 100 - 6gt = 0\). With \(g=10\text{ m/s}^2), (t = 100/60 = 5/3\text{ s}\). The assertion states \(4/3\text{ s})\, so (A) is False.
Reason (R): The dot product of two perpendicular vectors is indeed zero. So (R) is True.
Since (A) is false and (R) is true, none of the given options are strictly correct. However, if (A) is false, options (1), (2), (3) are ruled out, leaving (4) by elimination, despite (R) being true.
Assertion (A): In projectile motion (from ground to ground projection), horizontal range is always same for angle of projection \(\theta\) and \(90^\circ – \theta\).
Reason (R): Horizontal range is independent of angle of projection.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): The range formula is \(R = \frac{u^2 sin(2\theta)}{g}\). For angle \((90^\circ - \theta)\), \(R' = \frac{u^2 sin(2(90^\circ - \theta))}{g} = \frac{u^2 sin(180^\circ - 2\theta)}{g} = \frac{u^2 sin(2\theta)}{g} = R\). So (A) is True.
Reason (R): The horizontal range clearly depends on the angle of projection (theta) via (sin(2theta)). So (R) is False.
Since (A) is true and (R) is false, option (3) is correct.
A particle is projected with a speed of \( 20 \, \text{m s}^{-1} \) from level ground at an angle \( \theta \) equal to \( 45^\circ \) from horizontal. The ratio of maximum height attained by the body to horizontal range acquired by the body will be
1. \( \frac{2}{1} \)
2. \( \frac{1}{2} \)
3. \( \frac{1}{4} \)
4. \( \frac{3}{2} \)
View Answer
The ratio of maximum height \( H \) to horizontal range \( R \) is given by \( \frac{H}{R} = \frac{\tan \theta}{4} \). For \( \theta = 45^\circ \), \( \tan 45^\circ = 1 \), leading to \( \frac{H}{R} = \frac{1}{4} \).