Kinematics (Advanced pattern): questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 3 · AdvancedSingle correct
A particle is projected from level ground with speed 20m/s at 60∘ above the horizontal. Take g=10m/s2. What is the radius of curvature of its path at the instant (on the way up) when its velocity makes 30∘ with the horizontal?
A340m
B9803m
C380m
D3203m
Show answer and solution
Answer: B
Key insight: only the component of g perpendicular to the velocity bends the path, so R=v2/(gcosϕ), where ϕ is the angle of the velocity with the horizontal.
The horizontal component is conserved: vcos30∘=20cos60∘=10, so v=20/3 and v2=400/3.
Normal acceleration an=gcos30∘=53.
R=53400/3=3380=9803≈15.4m.
Traps: v2/g=40/3 ignores the cosϕ factor; 80/3 uses the launch speed; 203/3 multiplies by cosϕ instead of dividing.
Exam pattern: JEE Advanced – radius of curvature of a projectile using the normal component of g
Question 2Level 3 · AdvancedSingle correct
A rigid rod of length 5m has its lower end on a smooth horizontal floor and its upper end against a smooth vertical wall. The lower end is pulled away from the wall with a constant speed of 3m/s, and the upper end stays in contact with the wall. At the instant the lower end is 3m from the wall, what is the acceleration of the upper end?
AZero, because the lower end moves with constant velocity
B64125m/s2 downward
C64225m/s2 downward
D49m/s2 downward
Show answer and solution
Answer: C
Constraint: x2+y2=25. Differentiate once: xx˙+yy˙=0, so with x=3, y=4, x˙=3: y˙=−9/4m/s.
Differentiate again: x˙2+xx¨+y˙2+yy¨=0 with x¨=0.
y¨=−(x˙2+y˙2)/y=−(9+81/16)/4=−225/64m/s2.
The upper end accelerates downward at about 3.52m/s2 even though the lower end moves uniformly.
Trap: 9/4 is the speed, not the acceleration; "zero" assumes a uniformly moving end forces the other end to move uniformly.
Exam pattern: JEE Advanced – ladder/rod constraint with second-derivative of the length constraint
Question 3Level 2 · ExamSingle correct
Six particles start at the vertices of a regular hexagon of side a. Each moves with constant speed v and always heads directly towards the next particle in the anticlockwise order. After how much time do they meet?
A3v2a
Bv2a
Cva
D3v4a
Show answer and solution
Answer: B
Key insight: by symmetry the particles always form a shrinking regular hexagon, so look at the rate at which the gap between neighbours closes.
The chaser moves along the side with speed v; the chased particle moves at 60∘ (the exterior angle) to that side, so its component along the side, away from the chaser, is vcos60∘.
Rate of closing =v−vcos60∘=v/2, which stays constant.
Time =a/(v/2)=2a/v. Each particle also covers a path length 2a.
Traps: a/v is the square result; 2a/(3v) is the triangle result; 4a/(3v) uses the interior angle 120∘ with the wrong sign.
Exam pattern: JEE Advanced – symmetric pursuit of n particles on a regular polygon
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