Assertion (A): Path of a projected ball is parabolic in uniform gravitational field for oblique projection in absence of air resistance.
Reason (R): Gravitational force is always act perpendicular to velocity during the motion of a projectile.
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
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The path of a projectile is a parabola under constant gravitational acceleration. Gravitational force acts vertically downwards, which is perpendicular to the velocity only at the highest point of its trajectory, so R is false.
Assertion (A): Path of a projected ball is parabolic in uniform gravitational field for oblique projection in absence of air resistance.
Reason (R): Gravitational force is always act perpendicular to velocity during the motion of a projectile.
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
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Assertion (A) is true because projectile motion under gravity without air resistance follows a parabolic path.
Reason (R) is false as gravitational force acts perpendicular to velocity only at the peak of the trajectory, not always. Thus, (A) is true, (R) is false.
Assertion (A): Horizontal component of velocity is constant in projectile motion under gravity.
Reason (R): Two projectiles having same horizontal range must have the same time of flight.
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
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In projectile motion (neglecting air resistance), gravity acts only vertically. Thus, there is no horizontal acceleration, and the horizontal component of velocity remains constant. So (A) is true. Horizontal range is \(R = u_x T\). Projectiles launched at complementary angles have the same range but different times of flight (\(T = \frac{2u \sin\theta}{g}\)). So (R) is false.
Assertion (A): Trajectory of an object moving under a constant acceleration is a straight line.
Reason (R): The shape of trajectory depends only on the acceleration.
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
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An object under constant acceleration does not always follow a straight line (e.g., projectile motion is parabolic). A straight line occurs only if initial velocity is parallel or anti-parallel to acceleration. So (A) is false. The trajectory shape depends on both initial velocity and acceleration. So (R) is false.
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): 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
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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.
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.