Constraint Motion — String constraint, wedge constraint
Laws of Motion
9
JEE Qs
8%
Hard
75
min
Always draw clear, labelled diagrams, define a consistent coordinate system, and focus on establishing the geometric relationship that remains constant before differentiating.
🧮 Key Formulas
✅ Key Points for JEE
- 1Identify the constant length (for string) or constant geometric relationship (for wedge) that links the motion of different bodies.
- 2Express the total length of the string or the geometric relationship using position coordinates of the relevant bodies/points. Use trigonometry carefully.
- 3Differentiate the relationship with respect to time once to obtain velocity relations and twice to obtain acceleration relations. Pay attention to chain rule and implicit differentiation.
- 4The component of velocity of all points along a taut string is the same. For ideal pulleys, the string length over the pulley surface is often considered constant, simplifying relationships.
- 5For wedge constraints, the relative velocity of the two bodies along the common normal at the point of contact must be zero, meaning their velocities perpendicular to the contact surface are equal.
⚠️ Common Mistakes
- ✕Incorrectly defining the variable lengths or positions, leading to wrong constraint equations.
- ✕Errors in differentiation, especially when dealing with complex trigonometric relationships or implicit functions.
- ✕Confusing relative velocities perpendicular to the contact surface with other components, or misidentifying the common normal vector for wedge constraints.
- ✕Not consistently applying sign conventions for directions of velocities and accelerations, leading to algebraic errors.
📝 Practice Questions
See allQ41.A balloon and its content having mass M is moving up with an acceleration ' a '. The mass that must be released from the content so that the balloon starts moving up with an acceleration ' 3a′ will be (Take ' g ' as acceleration due to gravity) (1) 2Ma (2) 3Ma 3a+g 2a−g (3) 3Ma (4) 2Ma 2a+g 3a−g
Q33.A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e. dmdt ∝√v . If P is the power delivered to run the belt at constant speed then which of the following relationship is true? 2025 (29 Jan Shift 2) JEE Main Previous Year Paper (1) P ∝√v (2) P ∝v (3) P2 ∝v5 (4) P2 ∝v3
Q29.A car of mass ' m ' moves on a banked road having radius ' r ' and banking angle θ. To avoid slipping from banked road, the maximum permissible speed of the car is v0 . The coefficient of friction μ between the wheels of the car and the banked road is θ θ tan tan 0−rg (1) μ = v2o+rg (2) μ = v2 θ tan θ rg−v2 tan rg+v2o 0 θ θ tan v2o+rg tan 0−rg (4) μ = (3) μ = v2 θ tan θ rg+v2 tan rg−v2o 0
Q47. A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes ( π3 ) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is … … ML.
Q5. A wooden block, initially at rest on the ground, is pushed by a force which increases linearly with time t. Which of the following curve best describes acceleration of the block with time: (1) (2) (3) (4)
Q21.Two forces ¯F1 and ¯F2 are acting on a body. One force has magnitude thrice that of the other force and the 1 ) resultant of the two forces is equal to the force of larger magnitude. The angle between →F1 and →F2 is cos−1 ( n . The value of |n| is _____.
NCERT Chapters
- Class 11 Physics Ch 3: Motion in a Straight Line
- Class 11 Physics Ch 4: Motion in a Plane
- Class 11 Physics Ch 5: Laws of Motion