Centre of Mass, Momentum, Impulse and Collisions: 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 thin uniform circular disc has radius R and centre O. Take x- and y-axes in the plane of the disc with origin at O. Two circular holes, each of radius R/3, are cut out of the disc; their centres are at (R/2,0) and (0,R/2). (The holes lie entirely inside the disc and do not overlap.) The distance of the centre of mass of the remaining plate from O is ______. Exactly one option is correct.
AR/14
B2R/14
C2R/7
D2R/9
Show answer and solution
Answer: B
Treat each hole as a negative mass. With areal density σ, the full disc has mass M=σπR2 and each hole has mass σπ(R/3)2=M/9.
xcm=M−2M/9M⋅0−(M/9)(R/2)−(M/9)⋅0=7/9−R/18=−14R, and by symmetry ycm=−14R.
Distance from O: 2R/14, on the line y=x on the side away from the holes. Answer (B).
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
A uniform plank of mass 60kg and length 6m lies on a frictionless horizontal ice surface. A man of mass 40kg stands at the left end of the plank and a boy of mass 20kg stands at the right end; everything is initially at rest. The man and the boy walk along the plank (in any manner, possibly at different times) and exchange their positions, after which both stand still on the plank. Treat the man and the boy as particles. Relative to the ice, the plank finally ______. Exactly one option is correct.
Ais displaced 1m to the right
Bis displaced 2m to the left
Creturns to its original position, since only internal forces act
Dis displaced 1m to the left
Show answer and solution
Answer: D
No horizontal external force acts on man + boy + plank, and the system starts at rest, so the centre of mass stays fixed. Take x along the plank, origin at the initial left end. Let the plank finally be displaced by d (positive to the right).
Initially: 40(0)+20(6)+60(3)=300kg m.
Finally the man is at 6+d, the boy at d, the plank centre at 3+d: 40(6+d)+20d+60(3+d)=420+120d.
420+120d=300⇒d=−1m. The plank ends up displaced 1m to the left, whatever the manner of walking; once all are at rest relative to each other, total momentum zero forces everything to be at rest. Answer (D).
Difficulty: Hard
Question 3Level 3 · AdvancedSingle correct
A wedge of mass 3m has a smooth plane inclined face making 30∘ with the horizontal and rests on a smooth horizontal floor. A small block of mass m is released from rest on the inclined face at a height h above the bottom edge of the face. The block slides down without leaving the face and the wedge does not tilt. When the block reaches the bottom of the face, the speed of the wedge relative to the floor is ______. Exactly one option is correct.
A3gh/26
Bgh/8
C3gh/13
Dgh/26
Show answer and solution
Answer: A
Let the block's velocity relative to the wedge be w down the incline, and let the wedge move with speed V in the opposite horizontal direction.
Block velocity (ground): horizontal wcos30∘−V, vertical −wsin30∘.
Horizontal momentum (no horizontal external force, starts from zero): m(wcos30∘−V)=3mV⇒w=cos30∘4V=38V.
Energy (all surfaces smooth): 21m[(wcos30∘−V)2+w2sin230∘]+21(3m)V2=mgh.
Here wcos30∘−V=3V and wsin30∘=4V/3, so 21m(9V2+316V2)+23mV2=mgh, i.e. (643+69)V2=326V2=gh, giving V2=263gh.
(Check with the general result V2=(M+m)(M+msin2θ)2ghm2cos2θ=4(13/4)2gh(3/4)=263gh.) Answer (A).
Difficulty: Very hard
Get the full chapter for ₹99
Timed attempt with real exam marking, your rank among everyone who took it, solutions for every question, and spaced revision of your mistakes.