JEE Physics Collisions MCQs Set 04

Answer: (c) i, ii

Explanation:
1. Walking along the rails, the car also moves, so the total kinetic energy is (1/2)μv² with μ = mM/(m + M) < m.
2. Walking normal to the rails, the rails prevent the car from moving sideways, so the car stays at rest and the work is (1/2)mv².

Teacher's Note:
1. Work done equals the total kinetic energy created.
2. Use the reduced mass when the car can recoil.

Answer: (a) ii, iii, iv

Explanation:
1. Maximum compression x = 2F m_right/(k(m_left + m_right)).
2. Case 1: 4F/(3k); case 2: F/(2k); case 3: 3F/(2k).
3. So x2 < x1 < x3.

Teacher's Note:
1. At maximum compression the relative velocity is zero.
2. The equilibrium compression is F m2/(k(m1 + m2)); the maximum is twice that.

Answer: (a) ii, iv

Explanation:
1. After k blocks have merged the speed is v/k, so the time to reach the next block is kL/v.
2. Total time = (L/v)(1 + 2 + ... + (n - 1)) = n(n - 1)L/(2v).
3. Finally all n blocks move together with speed v/n.

Teacher's Note:
1. Momentum conservation gives speed v/k for k merged blocks.
2. Sum the series of times.

Answer: (b) i, ii, iv

Explanation:
1. e = 2/4 = 0.5.
2. 12 = 3v1 + 2v2 and v2 - v1 = 2 give v1 = 1.6 m/s, v2 = 3.6 m/s.
3. Impulse = 2 × 3.6 = 7.2 N s. KE loss = 24 - (3.84 + 12.96) = 7.2 J.

Teacher's Note:
1. Use e = separation speed/approach speed.
2. Impulse equals the change in momentum of either body.

Answer: (d) i, ii, iv

Explanation:
1. Let B = m and A = 2m. Momentum: mu = 3mv, so v = u/3.
2. Loss of KE = (1/2)mu² - (1/2)(3m)(u/3)² = (2/3)((1/2)mu²).
3. a_B = μg (decelerating) and a_A = μmg/2m = μg/2, so the relative acceleration is (3/2)μg.

Teacher's Note:
1. Use momentum conservation for the common velocity.
2. Relative acceleration is the sum of the magnitudes.

Answer: (d) i, ii, iii

Explanation:
1. The wall is along ĵ, so the x-component (normal) becomes -(1/2)(3) = -3/2 and the y-component stays 1.
2. KE loss = (1/2)m(9 - 9/4) = 27m/8.
3. Impulse = m(-3/2 - 3)î = -(9/2)m î.

Teacher's Note:
1. Only the component normal to the wall changes.
2. Impulse equals the change in momentum.

Answer: (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Explanation:
1. Elastic collisions have e = 1, so the relative speed is unchanged. Momentum is conserved in every collision.
2. The reason is true but it is not why the relative speed is unchanged; that comes from the conservation of kinetic energy as well.

Teacher's Note:
1. Relative speed reversal needs both momentum and energy conservation.
2. Momentum conservation alone is not enough.

Answer: (d) Assertion is false but Reason is true.

Explanation:
1. During contact some kinetic energy is stored as elastic potential energy, so the kinetic energy is not conserved at each instant; the assertion is false.
2. Energy spent against friction is converted into heat and total energy is conserved, so the reason is true.

Teacher's Note:
1. Kinetic energy is regained only after the deformation recovers.
2. Total energy is always conserved.

Answer: (c) \( \rho A(v-v_B)^2 \)

Explanation:
1. Speed of the jet relative to the blade = v - vB.
2. Mass hitting the blade per second = ρA(v - vB), and its x-momentum (relative to the blade) is destroyed.
3. Fx = ρA(v - vB)².

Teacher's Note:
1. Work in the frame of the moving blade.
2. Only the relative speed matters for the rate of mass striking the blade.

Answer: (c) 0.75 J

Explanation:
1. vA = 1 m/s from momentum conservation.
2. Stored energy = (1/2)(1)(1)² + (1/2)(2)(0.5)² = 0.5 + 0.25 = 0.75 J.

Teacher's Note:
1. All the stored energy becomes kinetic energy of the blocks.
2. Add the two kinetic energies.

Answer: (d) \( \sqrt{105} \) m/s

Explanation:
1. Drop along BC = 3√3 tan 30° = 3 m.
2. v² = 45 + 2(10)(3) = 105, so v = √105 m/s.

Teacher's Note:
1. Use the speed just after the impact at B as the initial speed.
2. Work-energy over the 3 m drop.

Answer: (c) 0

Explanation:
1. The velocity before impact makes 30° with the second incline (60° vs 30° below horizontal).
2. In an elastic collision the normal component reverses, so the velocity after impact makes 30° on the other side of the incline, which is horizontal.
3. So the vertical component is zero.

Teacher's Note:
1. Reflect the velocity about the incline surface.
2. The angles with the horizontal are -60° (before) and 0° (after).

Answer: (c) \( \tan^{-1}\sqrt e \)

Explanation:
1. Tangential component is unchanged: v cos θ = u sin θ.
2. Normal component: v sin θ = e u cos θ.
3. Dividing: tan θ = e cot θ, so tan²θ = e and θ = tan⁻¹√e.

Teacher's Note:
1. Resolve along and perpendicular to the plane.
2. The tangential component is unchanged.

Answer: (c) \( v=u\sqrt e \)

Explanation:
1. Tangential component: v cos θ = u sin θ, so v = u tan θ.
2. With tan θ = √e, v = u√e.

Teacher's Note:
1. Use the result tan²θ = e from the previous step.
2. Substitute into v = u tan θ.

Answer: (d) \( 2u\sin\alpha/g \)

Explanation:
1. The wall changes only the horizontal velocity; the vertical motion is unaffected.
2. The sphere returns to the ground level at O, so the total time is 2u sin α/g.

Teacher's Note:
1. Vertical motion is independent of the wall.
2. The return to the same level fixes the total time.

Answer: (b) \( \frac{25}{3} \) m/s

Explanation:
1. Block A acquires speed from the bullet: 1 × vA = 0.25(200 - 100), so vA = 25 m/s.
2. When the string becomes taut, A and B move with a common speed: 1(25) = (1 + 2)v, so v = 25/3 m/s.

Teacher's Note:
1. The string jerk gives the same speed to A and B.
2. Momentum along the string (taking the impulsive tension as internal) is conserved.

Answer: (a) A-r, B-p, C-q, D-s

Explanation:
1. Velocity of the first after collision = -u/3 and of the second = 2u/3.
2. Momentum of the first = -p/3; of the second = 2m(2u/3) = 4p/3.
3. KE of the first = K/9; of the second = (1/2)(2m)(4u²/9) = 8K/9.

Teacher's Note:
1. Use the elastic collision formulas for m and 2m.
2. Then compute momenta and kinetic energies.

Answer: (b) \( \frac{2}{\sqrt3}v \)

Explanation:
1. Momentum along the initial direction: mv = m v2x, so v2x = v.
2. Perpendicular direction: 0 = m(v/√3) + m v2y, so v2y = -v/√3.
3. Speed = √(v² + v²/3) = 2v/√3.

Teacher's Note:
1. Conserve momentum in two perpendicular directions.
2. Combine the components of the second mass.

Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Explanation:
1. After a completely inelastic collision the combined mass still moves (momentum cannot vanish), so some kinetic energy remains.
2. This follows from momentum conservation, so the reason explains the assertion.

Teacher's Note:
1. A non-zero total momentum needs a non-zero velocity.
2. Momentum is conserved in every collision.

Answer: (c) 2

Explanation:
1. Equal masses in an elastic collision exchange velocities, which is equivalent to the particles passing through each other.
2. Relative speed = 3v. The faster particle returns to A at time 2πR/v (two revolutions) when the slower has made one revolution.
3. Distance covered relative = 3v × 2πR/v = 3 circumferences, so there are 3 meetings, the last one at A. Collisions other than at A = 2.

Teacher's Note:
1. Treat the exchange of velocities as ghost particles passing through.
2. Count the meetings in the time when both are back at A.

Answer: (b) \( \frac a{10} \)

Explanation:
1. y_cm = [6m(0) + m(a) + m(a) + m(0) + m(-a)]/(10m) = a/10.

Teacher's Note:
1. Add up m × y for every part.
2. The total mass is 6m + 4m = 10m.

Answer: (c) 500 m/s

Explanation:
1. Fall time = √(2 × 5/10) = 1 s.
2. Ball speed = 20 m/s and bullet speed = 100 m/s after the collision.
3. Momentum: 0.01V = 0.2(20) + 0.01(100) = 5, so V = 500 m/s.

Teacher's Note:
1. Both fall through 5 m in the same time.
2. Horizontal speeds follow from the distances.

Answer: (d) Assertion is false but Reason is true.

Explanation:
1. The maximum loss fraction is M/(M + m), not m/(M + m), so the assertion is false.
2. The maximum loss indeed occurs in the perfectly inelastic collision, so the reason is true.

Teacher's Note:
1. Loss = (1/2)μv² with μ = mM/(m + M), so the fraction is M/(m + M).
2. Perfectly inelastic collisions lose the most energy.