Assertion (A): If a body moves on a straight line, magnitude of its displacement and distance covered by it must be same.
Reason (R): Along a straight line, a body can move only in one direction.
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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Solution: (A) is false; if a body moves forward and then reverses on a straight line, distance will be greater than magnitude of displacement. (R) is false; a body can change its direction of motion while staying on a straight line (e.g., moving forward, then backward). Since both are false, option (4) is correct.
Assertion (A): An object moving with a velocity of magnitude \(10 \text{ m/s}\) is subjected to a uniform acceleration \(2 \text{ m/s}^2\) at right angle to the initial motion. Its velocity after \(5s\) has a magnitude nearly \(14 \text{ m/s}\).
Reason (R): The equation \(\vec{v} = \vec{u} + \vec{a}t\) can be applied to obtain \(\vec{v}\) if \(\vec{a}\) is constant.
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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Assertion (A): Given \(u = 10 \text{ m/s}\), \(a = 2 \text{ m/s}^2\), \(t = 5 \text{ s}\). Since \(\vec{u}\) and \(\vec{a}\) are perpendicular, the final velocity magnitude is \(|\vec{v}| = \sqrt{u^2 + (at)^2} = \sqrt{10^2 + (2 \times 5)^2} = \sqrt{100+100} = \sqrt{200} \approx 14.14 \text{ m/s}\). So (A) is true.
Reason (R): The equation \(\vec{v} = \vec{u} + \vec{a}t\) is valid when acceleration \(\vec{a}\) is constant. So (R) is true.
(R) correctly explains (A) as the formula is used due to constant acceleration.
Assertion (A): A coin is allowed to fall in a train moving with constant velocity. Its trajectory is a straight line as seen by observer attached to the train.
Reason (R): An observer on ground will see the path of coin as a parabola.
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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Assertion (A): From the train's frame of reference (inertial, moving with constant velocity), the coin only has vertical motion under gravity, thus appearing as a straight line. So (A) is true.
Reason (R): From the ground frame, the coin has an initial horizontal velocity (that of the train) and vertical acceleration due to gravity, resulting in a parabolic path. So (R) is true.
However, (R) describes a different frame of reference and does not explain why the path is a straight line in the train's frame.
Assertion (A): Two particles start moving with velocities \(vec{v}_1\) and \(vec{v}_2\) respectively in a plane. They can meet only if component of their velocities perpendicular to line joining them are equal.
Reason (R): Relative velocity of a body w.r.t. other body is calculated along the line joining two bodies.
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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Assertion (A): For particles to meet, their relative perpendicular velocity component must be zero, meaning their perpendicular velocities must be equal. Otherwise, they would move apart perpendicular to the line joining them. So (A) is true.
Reason (R): Relative velocity is a vector difference and can be calculated in any direction, not exclusively along the line joining two bodies. So (R) is false.
Assertion (A): Two balls are dropped one after the other from a tall tower. The distance between them increases linearly with time (elapsed after the second ball is dropped and before the first hits ground).
Reason (R): In given situation relative acceleration is zero, whereas relative velocity is non-zero.
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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Assertion (A): Let \(\Delta t\) be the time interval. The distance between them \(D(t) = \frac{1}{2}gt^2 - \frac{1}{2}g(t-\Delta t)^2 = g t \Delta t - \frac{1}{2}g (\Delta t)^2\). This is a linear function of time \(t\). So (A) is true.
Reason (R): Both balls accelerate at \(g\). Thus, their relative acceleration is \(\vec{g} - \vec{g} = \vec{0}\). The first ball has velocity \(g\Delta t\) when the second is dropped, so the relative velocity is non-zero and constant. So (R) is true.
(R) correctly explains (A): constant non-zero relative velocity results in linear increase in relative distance.
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
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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
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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
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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.