NEET UG Question of the Day

The horizontal range (R) of a projectile launched with an initial speed 'u' at an angle 'θ' with the horizontal is given by the formula:

R = (u^2 * sin(2θ)) / g

Given values are: Initial speed (u) = 10 m/s, launch angle (θ) = 30 degrees, and acceleration due to gravity (g) = 10 m/s^2.

Substitute these values into the range formula:

R = (10^2 * sin(2 * 30°)) / 10

R = (100 * sin(60°)) / 10

We know that sin(60°) = sqrt(3)/2 ≈ 0.866.

R = (100 * 0.866) / 10

R = 10 * 0.866

R = 8.66 meters

Rounding this to one decimal place, the approximate maximum horizontal distance covered is 8.7 meters.

Distractor Analysis:

• A value of 5.0 meters would be obtained if the student incorrectly used sin(θ) instead of sin(2θ) in the range formula. This would lead to R = (100 * sin(30°)) / 10 = (100 * 0.5) / 10 = 5 meters.
• A value of 1.25 meters represents the maximum height (H) reached by the droplet, calculated using H = (u^2 * sin^2(θ)) / (2g) = (100 * (sin(30°))^2) / (2 * 10) = (100 * (0.5)^2) / 20 = (100 * 0.25) / 20 = 25 / 20 = 1.25 meters. This is a common confusion between range and maximum height.
• A value of 10.0 meters would result if the student incorrectly assumed sin(2θ) = 1 (which only occurs for a launch angle of 45 degrees), leading to R = u^2/g = 10^2/10 = 10 meters.

According to the principle of homogeneity of dimensions, every term in a physical equation must have the same dimensions. In the given equation, F = Av + Bv^2, each term (F, Av, and Bv^2) must have the dimensions of force.

The dimensional formula for Force (F) is [MLT^-2].
The dimensional formula for Velocity (v) is [LT^-1].

For the term Av to have the dimensions of Force:
[A] * [v] = [F]
[A] * [LT^-1] = [MLT^-2]
[A] = [MLT^-2] / [LT^-1]
[A] = [M L^(1-1) T^(-2 - (-1))]
[A] = [M L^0 T^-1] = [MT^-1]

For the term Bv^2 to have the dimensions of Force:
[B] * [v^2] = [F]
[B] * ([LT^-1]^2) = [MLT^-2]
[B] * [L^2T^-2] = [MLT^-2]
[B] = [MLT^-2] / [L^2T^-2]
[B] = [M L^(1-2) T^(-2 - (-2))]
[B] = [M L^-1 T^0] = [ML^-1]

Thus, the correct dimensional formulae for A and B are [MT^-1] and [ML^-1] respectively.

The option 'MLT^-1, ML^-1' is incorrect for constant A, as it suggests an error where the length dimension 'L' is not correctly cancelled during the calculation for A, although the dimension for B is correctly identified. The option 'MT^-2, ML^-1T^-2' shows errors in calculating both A and B, possibly from confusing them with the dimension of force or incorrect exponent manipulation. The option 'ML^2T^-1, ML^-2' reflects significant calculation errors in determining the powers of L and T for both constants, such as improper squaring of velocity's dimensions or incorrect division of terms.

To determine the friction force acting on block A, we first need to check if the blocks move together or if block A slips relative to block B. The maximum static friction force that can act on block A is f_s_max = μ_s * N_A, where N_A is the normal force on A. Since A is on a horizontal surface, N_A = m_A * g. \nTherefore, f_s_max = 0.8 * 1 kg * 10 m/s^2 = 8 N.

If the blocks move together, they share a common acceleration 'a'. The total mass of the system is M_total = m_A + m_B = 1 kg + 2 kg = 3 kg. The applied force F = 15 N. Since the bottom surface is smooth, there is no friction from the ground. So, the common acceleration a = F / M_total = 15 N / 3 kg = 5 m/s^2.\nFor block A to move with this acceleration 'a', the static friction force from B on A must provide this acceleration. So, f_static_on_A = m_A * a = 1 kg * 5 m/s^2 = 5 N.\nSince this required static friction (5 N) is less than the maximum possible static friction (8 N), the blocks indeed move together without slipping. Thus, the magnitude of the friction force acting on block A is 5 N.

Let's analyze the distractors:\n* 8 N: This is the maximum possible static friction force (f_s_max = μ_s * m_A * g), not the actual static friction required for the given acceleration. The actual static friction adjusts itself to provide the necessary acceleration, up to this maximum value.\n* 6 N: This is the kinetic friction force (f_k = μ_k * m_A * g = 0.6 * 1 kg * 10 m/s^2 = 6 N) that would act on block A if it were slipping relative to block B. However, in this scenario, the blocks move together, so static friction is at play.\n* 18 N: This value could arise from an incorrect application of kinetic friction to the total mass (μ_k * (m_A + m_B) * g = 0.6 * (1+2) kg * 10 m/s^2 = 18 N), which is irrelevant for the friction force on block A alone when the blocks move together. It represents a common conceptual mix-up regarding which mass and which friction coefficient to apply in a multi-block system.

The generalized work-energy theorem states that the work done by all non-conservative forces (W_nc) acting on a system is equal to the change in its mechanical energy (ΔE_mechanical). This can be expressed as W_nc = ΔE_mechanical = ΔKE + ΔPE.

In this problem, the system is the block. The conservative forces acting on the block are gravity and the normal force. Since the motion is strictly horizontal, the work done by gravity and the normal force is zero. The non-conservative forces are the applied force 'F' and the kinetic friction force 'f_k'.

1. **Work done by the applied force (F):** Since the force is constant and in the direction of displacement, the work done by F is W_F = Fd.

2. **Work done by kinetic friction (f_k):** The kinetic friction force opposes the motion. Its magnitude is f_k = μN. On a horizontal surface, the normal force N equals the gravitational force mg. So, f_k = μmg. The work done by friction is W_friction = -f_k * d = -μmgd. The negative sign indicates that the friction force opposes the displacement.

3. **Net work done by non-conservative forces (W_nc):** This is the sum of the work done by the applied force and the friction force. W_nc = W_F + W_friction = Fd - μmgd = (F - μmg)d.

4. **Change in mechanical energy (ΔE_mechanical):** According to the generalized work-energy theorem, ΔE_mechanical = W_nc. Therefore, ΔE_mechanical = (F - μmg)d. Since the motion is horizontal, the change in potential energy (ΔPE) is zero, so ΔE_mechanical = ΔKE + 0 = ΔKE. This means the change in kinetic energy is also (F - μmg)d.

Based on this analysis:

• The statement 'Work done by net non-conservative forces = (F - μmg)d; Change in mechanical energy = (F - μmg)d' accurately reflects both the calculated net non-conservative work and its direct relationship to the change in mechanical energy.
• The statement 'Work done by net non-conservative forces = -μmgd; Change in mechanical energy = (F - μmg)d' is incorrect because the applied force 'F' is also a non-conservative force in this context (as it's not derived from a potential energy of the system itself) and its work (Fd) must be included in the net non-conservative work.
• The statement 'Work done by net non-conservative forces = (F - μmg)d; Change in mechanical energy = 0' is incorrect. Mechanical energy is not conserved when non-conservative forces do a net non-zero work (assuming F ≠ μmg), which is the case here. The change in mechanical energy must be equal to the net non-conservative work.
• The statement 'Work done by net non-conservative forces = (F + μmg)d; Change in mechanical energy = (F + μmg)d' is incorrect because the work done by kinetic friction is always negative as it opposes motion, not positive.

Assertion (A) is true. According to Newton's second law for a system of particles, the net external force (F_ext) acting on the system is equal to the product of the total mass (M) of the system and the acceleration of its center of mass (a_CM), i.e., F_ext = M * a_CM. If the net external force (F_ext) is zero, then the acceleration of the center of mass (a_CM) must be zero. Since acceleration is the rate of change of velocity, a zero acceleration implies that the velocity of the center of mass (V_CM) is constant.

Reason (R) is also true. The principle of conservation of linear momentum states that if the net external force acting on a system is zero (i.e., an isolated system), the total linear momentum of the system remains constant. This is a fundamental principle derived from Newton's laws. Furthermore, Reason (R) is the correct explanation for Assertion (A). The total linear momentum (P_total) of a system of particles is defined as the product of the total mass (M) of the system and the velocity of its center of mass (V_CM), i.e., P_total = M * V_CM. If, as stated in Reason (R), the total linear momentum (P_total) is conserved (constant) for an isolated system, and the total mass (M) of the system is constant (assuming no mass exchange with the surroundings), then the velocity of the center of mass (V_CM) must necessarily be constant. Thus, the conservation of total linear momentum directly explains why the velocity of the center of mass remains constant when the net external force is zero.