JEE Main Question of the Day

Let the position vector of particle P at time `t` be `vec(r_P)(t)` and that of particle Q be `vec(r_Q)(t)`. The acceleration of particle P is `vec(a_P) = vec(g)` and the acceleration of particle Q is `vec(a_Q) = vec(g)`. Both particles experience the same constant acceleration due to gravity.

The acceleration of particle Q as observed from particle P (relative acceleration) is given by `vec(a_{QP}) = vec(a_Q) - vec(a_P)`. Since `vec(a_Q) = vec(g)` and `vec(a_P) = vec(g)`, we have `vec(a_{QP}) = vec(g) - vec(g) = vec(0)`. This means that the relative acceleration between the two particles is zero.

If the relative acceleration is zero, the relative velocity `vec(v_{QP}) = vec(v_Q) - vec(v_P)` must be constant. A body moving with a constant velocity follows a straight-line path. Therefore, the path of particle Q as observed from particle P will be a straight line. Since both particles start from the same point, the relative position vector `vec(r_{QP}) = vec(r_Q) - vec(r_P)` starts from `vec(0)` at `t=0` and evolves as `vec(r_{QP}) = vec(v_{QP}) t`, which is the equation of a straight line passing through the origin.

Why other options are incorrect:

• A parabolic path: This would be the path of an individual projectile observed from a stationary frame of reference (like the ground), because its acceleration is constant and non-zero (equal to `vec(g)`). However, for relative motion between two projectiles under the same `vec(g)`, the relative acceleration is zero, leading to a straight path.
• A circular path: Circular motion requires a continuous change in the direction of velocity due to a centripetal acceleration, which is not constant or zero in this scenario. Projectile motion, whether absolute or relative (if relative acceleration is non-zero), does not result in a circular path.
• A path whose shape depends on the magnitudes and directions of `vec(u_P)` and `vec(u_Q)`: While the magnitude and direction of the constant relative velocity `vec(v_{QP})` (and thus the slope and speed of the straight line) depend on `vec(u_P)` and `vec(u_Q)`, the *shape* of the relative path itself (i.e., being a straight line) is determined solely by the fact that the relative acceleration is zero, irrespective of the initial velocity vectors.

First, let's analyze the initial state of the system to confirm it is indeed at rest.

The force tending to move the system is the tension (T) created by the hanging mass m. In equilibrium, T = mg = 2 kg * 10 m/s² = 20 N.

The maximum static friction (f_s,max) that can oppose the motion of block M is given by f_s,max = μ_s * N, where N is the normal force acting on block M. Since M is on a horizontal surface, N = Mg. So, f_s,max = μ_s * Mg = 0.5 * 5 kg * 10 m/s² = 25 N.

Since the initial tension (20 N) is less than the maximum static friction (25 N), the system is initially at rest, as stated in the problem.

For the system to just begin to move, the new tension (T') must be equal to the maximum static friction. Let m_add be the additional mass added to the hanging block m. The total hanging mass will then be (m + m_add).

The new tension T' = (m + m_add)g.

Setting T' equal to f_s,max:
(m + m_add)g = μ_sMg
(2 + m_add) * 10 = 0.5 * 5 * 10
(2 + m_add) * 10 = 25
2 + m_add = 25 / 10
2 + m_add = 2.5
m_add = 2.5 - 2
m_add = 0.5 kg.

Therefore, the minimum additional mass that must be added to the hanging block is 0.5 kg.

Let's analyze the distractors:

• 2.5 kg: This value represents the total mass (m + m_add) that should be hanging for the system to just begin to move (i.e., μ_sM = 2.5 kg). A student might incorrectly state this as the additional mass, forgetting to subtract the initial mass m.
• 1.5 kg: This distractor could arise from a conceptual error where a student might incorrectly assume the friction acts on a total mass of (M + m) instead of just M. In that case, f_s,max = μ_s(M+m)g = 0.5 * (5+2) * 10 = 35 N. Then (m + m_add)g = 35, so 2 + m_add = 3.5, giving m_add = 1.5 kg.
• 4.5 kg: This could be a result of adding the initial hanging mass (m) to the value of μ_sM (2.5 kg), which is `2.5 + 2 = 4.5 kg`. This suggests a misunderstanding of how the forces balance.

First, we analyze the elastic collision between the two blocks. Let m1 = 2 kg, u1 = 10 m/s, m2 = 3 kg, and u2 = 0 m/s. For a head-on elastic collision, the final velocities are given by:

v1_f = ((m1 - m2) / (m1 + m2)) * u1 + (2 * m2 / (m1 + m2)) * u2
v2_f = ((2 * m1) / (m1 + m2)) * u1 + ((m2 - m1) / (m1 + m2)) * u2

Since u2 = 0, these simplify to:
v1_f = ((2 - 3) / (2 + 3)) * 10 = (-1/5) * 10 = -2 m/s (The 2 kg block reverses direction, as stated in the problem).
v2_f = ((2 * 2) / (2 + 3)) * 10 = (4/5) * 10 = 8 m/s (The 3 kg block moves forward).

Next, the 3 kg block (m2) moves with a velocity of v2_f = 8 m/s and compresses the spring. The kinetic energy of the 3 kg block is converted into the potential energy stored in the spring when the compression is maximum. At maximum compression 'x', the velocity of the 3 kg block becomes zero.

Using the principle of conservation of energy:
(1/2) * m2 * v2_f^2 = (1/2) * k * x^2
m2 * v2_f^2 = k * x^2
3 kg * (8 m/s)^2 = 1500 N/m * x^2
3 * 64 = 1500 * x^2
192 = 1500 * x^2
x^2 = 192 / 1500 = 64 / 500 = 16 / 125
x = sqrt(16 / 125) = 4 / sqrt(125) = 4 / (5 * sqrt(5))
x = (4 * sqrt(5)) / 25

Using sqrt(5) approx 2.236:
x = (4 * 2.236) / 25 = 8.944 / 25 = 0.35776 m

Rounding to three significant figures, the maximum compression x = 0.358 m.

Rationale for distractors:
- A value of 0.231 m would be obtained if the collision were incorrectly assumed to be perfectly inelastic (blocks stick together). In that case, the combined velocity would be ((2 * 10) / (2 + 3)) = 4 m/s. Then, (1/2) * (5) * (4)^2 = (1/2) * 1500 * x^2, leading to x = sqrt(80/1500) = sqrt(4/75) approx 0.231 m.
- A value of 0.447 m would be obtained if it was incorrectly assumed that the 3 kg block (m2) acquired the initial velocity of the 2 kg block (u1=10 m/s). Then (1/2) * 3 * (10)^2 = (1/2) * 1500 * x^2, leading to x = sqrt(300/1500) = sqrt(1/5) approx 0.447 m.
- A value of 0.052 m would be obtained if the kinetic energy of the 2 kg block *after* the collision was incorrectly used to compress the spring. The 2 kg block moves with v1_f = -2 m/s. If its kinetic energy was used, (1/2) * 2 * (-2)^2 = (1/2) * 1500 * x^2, leading to x = sqrt(4/1500) = sqrt(1/375) approx 0.052 m.

The total mechanical energy (E) of the particle is conserved because no non-conservative forces are acting. The particle is released from rest at x = 0 m. At this initial position:

• Potential energy U(0) = 10 J (given).
• Kinetic energy K(0) = 0 J (since it's released from rest).

Therefore, the total mechanical energy E = U(0) + K(0) = 10 J + 0 J = 10 J.

For the particle to access any position x, its potential energy U(x) at that position must be less than or equal to its total mechanical energy E. In this problem, E = 10 J. By examining the given points, the potential energy values are 10 J, 0 J, 5 J, -5 J, and 10 J. All these potential energy values are less than or equal to 10 J, meaning the particle can potentially reach all these points. The maximum speed of the particle occurs when its kinetic energy is maximum. Kinetic energy is maximum when the potential energy is at its minimum value within the accessible region.

The minimum potential energy given in the graph is U_min = -5 J, which occurs at x = 6 m. At this point, the kinetic energy will be maximum:

K_max = E - U_min = 10 J - (-5 J) = 15 J.

Using the formula for kinetic energy, K_max = (1/2)mv_max^2:

15 J = (1/2)(0.5 kg)v_max^2

15 = (1/4)v_max^2

v_max^2 = 15 * 4 = 60

v_max = sqrt(60) = sqrt(4 * 15) = 2 * sqrt(15) m/s.

Distractor Analysis:

• 2 * sqrt(10) m/s: This result would be obtained if a student incorrectly assumes the minimum potential energy accessible is 0 J (e.g., at x=2m), perhaps by overlooking the negative potential energy values. In this case, K_max = 10 J - 0 J = 10 J, leading to v_max = sqrt(10 / 0.25) = sqrt(40) = 2 * sqrt(10) m/s.
• 2 * sqrt(5) m/s: This can arise from a sign error in calculating K_max, such as using K_max = E + U_min = 10 J + (-5 J) = 5 J. This would result in v_max = sqrt(5 / 0.25) = sqrt(20) = 2 * sqrt(5) m/s.
• sqrt(15) m/s: This is a result of an algebraic error, typically by omitting the (1/2)m factor or incorrectly manipulating it, effectively setting v_max^2 = K_max or assuming m=2kg when it's 0.5kg. For example, if K_max = 15 J and a student calculates v_max = sqrt(K_max) instead of sqrt(2*K_max/m).

The formula for the acceleration due to gravity is given by g = 4π²L/T².

For a physical quantity P expressed as a product or quotient of measured quantities with powers, P = (a^x * b^y) / c^z, the maximum fractional (or relative) error is given by the sum of the absolute values of the fractional errors of individual quantities multiplied by their respective powers:

ΔP/P = x(Δa/a) + y(Δb/b) + z(Δc/c).

In this problem, g = (4π²) * L^1 * T^(-2). The constant 4π² has no associated error.

So, the fractional error in g is:

Δg/g = (ΔL/L) + 2(ΔT/T)

Given values and their absolute errors:

• Length L = 100.0 cm, absolute error ΔL = 0.1 cm
• Time period T = 2.0 s, absolute error ΔT = 0.01 s

Calculate the relative error in L:

ΔL/L = 0.1 / 100.0 = 0.001

Calculate the relative error in T:

ΔT/T = 0.01 / 2.0 = 0.005

Substitute these values into the error propagation formula:

Δg/g = 0.001 + 2 * (0.005)

Δg/g = 0.001 + 0.010

Δg/g = 0.011

To find the percentage error, multiply the fractional error by 100%:

Percentage error in g = (Δg/g) * 100% = 0.011 * 100% = 1.1 %

Distractor analysis:

• A common mistake is to forget to multiply the relative error in T by its power (2). In this case, the calculation would be Δg/g = (ΔL/L) + (ΔT/T) = 0.001 + 0.005 = 0.006, leading to a percentage error of 0.6 %.
• Another common error is to incorrectly use a power of 1/2 for T (e.g., if one mistakenly thought g was proportional to sqrt(T)). This would lead to Δg/g = (ΔL/L) + (1/2)(ΔT/T) = 0.001 + 0.5 * 0.005 = 0.001 + 0.0025 = 0.0035, resulting in a percentage error of 0.35 %.
• A student might mistakenly apply the coefficient '2' to both terms, calculating Δg/g = 2(ΔL/L) + 2(ΔT/T) = 2 * 0.001 + 2 * 0.005 = 0.002 + 0.010 = 0.012, which leads to a percentage error of 1.2 %.