Kinematics in One and Two Dimensions: 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 2 · ExamSingle correct
A particle is projected from a point on level ground and moves under gravity alone. Take g=10m/s2 and neglect air resistance. Its speed at the highest point of its path is 10m/s, and its speed at the instant (on the way up) when it is at half of its maximum height is 510m/s. The maximum height reached above the ground is ______. Exactly one option is correct.
A10m
B15m
C20m
D7.5m
Show answer and solution
Answer: B
The horizontal velocity component is constant, so ucosθ=10m/s (speed at the top). At height H/2 the vertical component satisfies vy2=u2sin2θ−2g(H/2)=u2sin2θ−21u2sin2θ=21u2sin2θ, using H=u2sin2θ/(2g). Hence 100+21u2sin2θ=250, so u2sin2θ=300m2/s2 and H=300/20=15m. (Then tanθ=3, θ=60∘, u=20m/s.) Answer (B).
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
Two long, straight, thin rods lie one above the other on a smooth horizontal table, taken as the xy-plane. Rod 1 always stays parallel to the x-axis and translates with velocity 2m/s along +y. Rod 2 always makes an angle of 60∘ (measured anticlockwise) with the +x-axis and translates, perpendicular to its own length, with speed 2m/s in the direction that makes 30∘ below the +x-axis (that is, along the unit vector 23i^−21j^). Neither rod rotates. The speed of the point where the two rods cross is ______. Exactly one option is correct.
A2m/s
B23m/s
C34m/s
D4m/s
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Answer: D
Let n^1=j^ and n^2=23i^−21j^ be the directions of motion (each normal to its rod). A line moving perpendicular to itself with speed v is n^⋅r=vt+const. The crossing point P lies on both lines, so its velocity vP satisfies n^1⋅vP=2 and n^2⋅vP=2. Writing vP=(a,b): b=2 and 23a−1=2, so a=23. Hence vP=12+4=4m/s, directed at 30∘ above +x, along the bisector of n^1 and n^2. In general, if the normals make angle β, vP=v/cos(β/2); here β=120∘, giving 2v. Trap: the crossing point is not a material point, so its velocity is not the vector sum of the rods' velocities (that sum has magnitude only 2m/s, option A). Answer (D).
Difficulty: Very hard
Question 3Level 2 · ExamSingle correct
A long fixed plane is inclined at 30∘ to the horizontal. A ball is projected horizontally with speed 10m/s from a point on the plane, in the vertical plane containing the line of greatest slope and towards the downhill side, so that it leaves the plane and later lands on it. Take g=10m/s2 and neglect air resistance. The greatest perpendicular distance of the ball from the plane during this flight is ______. Exactly one option is correct.
A653m
B45m
C353m
D35m
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Answer: A
Take axes along the plane (down the slope) and along the normal to the plane (away from it). The launch velocity, horizontal, makes 30∘ with the plane, so its normal component is un=10sin30∘=5m/s away from the plane. The normal component of gravity is gcos30∘=53m/s2 towards the plane. The normal motion is a uniformly decelerated motion, so the greatest distance is 2gcos30∘un2=10325=653m≈1.44m, reached at t=un/(gcos30∘)=1/3s, exactly half the time of flight 2/3s. At that instant the velocity is parallel to the plane. (Using g instead of gcos30∘ gives B; doubling gives C.) Answer (A).
Difficulty: Hard
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