Module 1 — Problem Set 2

Before you start

Do Set 1 first. These assume you can already resolve a vector, take a derivative and evaluate an integral without stopping to think about it, because none of that is what these problems are testing.

Again there are no answers, and again each problem names the Foundations it draws on — several draw on more than one, which is the point. A JEE Advanced problem is set on a situation, not on a topic.

Three of these have a trap that a correct method will walk straight into. Watch for the moment your answer stops being about the physical situation you started with.

Two vectors A and B have equal magnitudes, A = B. Their sum has magnitude A as well.

(a) Find the angle between A and B.

(b) Suppose their difference also had magnitude A. Decide whether |A + B| = A and |A − B| = A can hold at the same time, and prove your answer with vector algebra rather than by trying values.

Two vectors satisfy |A| = 5, |B| = 8 and A · B = 20.

(a) Find the angle between them.

(b) A non-zero vector C of magnitude 6 satisfies A · C = 0 and B · C = 0. Explain, both geometrically and algebraically, why such a vector can exist in three dimensions, give a formula for its direction, and state the condition on A and B under which it could not exist.

A force F = (3î + 4ĵ − 2k̂) N acts at the point r₀ = (î − ĵ + 2k̂) m.

(a) Find the torque about the origin.

(b) Find the torque about the point A = (2, 1, −1) m.

(c) Find every point about which this force produces zero torque, and describe that set geometrically. What does the result tell you about moving a force along its own line?

A particle moves along a line with x(t) = t⁴ − 8t³ + 18t² metres, for t ≥ 0.

(a) Find every time in 0 < t < 6 at which the velocity is zero.

(b) For each of those times, examine the sign of the velocity just before and just after, and decide whether the particle actually reverses direction or merely pauses and continues the same way. Do not assume that zero velocity means a turning point.

A particle moves along the curve y = 4x², with its horizontal velocity held constant at dx/dt = 3 m/s.

(a) Find the vertical component of velocity as a function of x.

(b) Find the speed at x = 1 m.

(c) Find the acceleration vector at any point on the path. Comment on what you get, and on which familiar motion this resembles.

A particle follows the spiral r(θ) = R₀e^kθ with R₀ = 2 m and k = 0.5 rad⁻¹, while its angle grows as θ(t) = 2t rad.

(a) Find r and θ at t = 0.5 s.

(b) Starting from r̂ = cos θ î + sin θ ĵ, show that dr̂/dt = θ̇ θ̂.

(c) Differentiate r = r r̂ to obtain the velocity in polar components, and find the speed at t = 0.5 s. Say which term would vanish for a circular path, and why.

A particle has acceleration a(t) = 6t − t² m/s² over 0 ≤ t ≤ 6 s, and its velocity at t = 0 is −8 m/s.

(a) Sketch a(t) and find where it peaks.

(b) Find v(t), and state when the velocity is decreasing. Note that this is not the same question as when the particle is slowing down.

(c) Find every time in the interval at which the particle changes direction. A calculator will be needed.

(d) Find the net displacement and the total distance over the six seconds.

A particle on the x-axis has acceleration a = −kx³ with k > 0. At x = 0 its velocity is v₀, in the positive direction.

(a) No time appears anywhere in this problem. Set up an equation relating v and x directly, and say why integrating a with respect to t is not available to you here.

(b) Integrate to find v as an explicit function of x.

(c) Find the furthest point the particle reaches before stopping instantaneously.

A particle moves along r(t) = 3cos t î + 3sin t ĵ + 4t k̂ metres.

(a) Find its positions at t = 0 and t = π, and the displacement between them.

(b) Find the speed, and hence the path length travelled over the same interval. The speed is simpler than you might expect.

(c) Find the ratio of path length to displacement magnitude, and explain why this ratio must exceed 1 for any path that is not a straight line.

The acceleration–time graph of a particle is a single straight line from a = 4 m/s² at t = 0 to a = −2 m/s² at t = 6 s. Its initial velocity is 3 m/s.

(a) Find the velocity at t = 6 s using areas alone, without writing down a(t).

(b) Find the instant of maximum velocity and its value. Say which feature of the graph told you where to look.

The relativistic kinetic energy of a particle is K = mc²[(1 − v²/c²)^−1/2 − 1]. Use the expansion (1 − x)^−1/2 = 1 + ½x + (3/8)x² + …

(a) Expand K as far as the term in v⁴.

(b) Identify the classical kinetic energy inside your expansion, and say what the next term represents.

(c) Find the fractional error in using the classical formula at v = 0.1c.

A particle moves in the xy-plane with r(t) = (2t − t²) î + 3t ĵ metres, for 0 ≤ t ≤ 4.

(a) Give the position at t = 0, 1, 2, 3, 4, and sketch the path.

(b) Find v(t) and a(t), and the speed at t = 1 s and t = 3 s.

(c) Determine whether the particle is ever instantaneously at rest. Be careful: one component of the velocity does vanish.

(d) Find the net displacement over the four seconds and its magnitude, and say why this is not the distance travelled.

(e) Evaluate v(2) · a(2). Are the two perpendicular at that moment?

(f) Evaluate r(2) × v(2) and interpret its direction.

(g) List which Foundation each of the preceding parts drew on.

When you have finished

Three of these punish a correct method applied without thought. In one, a repeated root means zero velocity without a turning point. In another, a vanishing component of velocity is not the same as a vanishing velocity. In the third, a question about position cannot be answered by integrating with respect to time. If you did not meet all three, go back and find them.

Problem 12 asks you to name the Foundation behind each part. That is not a formality. If you can do it, you have the map of Module 1 in your head, which is what the module was for.

Covers: Foundations 2–10, most problems drawing on several  ·  Type: Problems, no answers  ·  Difficulty: Advanced