JEE Physics Laws Of Motion Notes Set 01

Newton's Laws of Motion

1. First law (inertia): a body stays at rest or in uniform motion along a straight line unless a net external force acts on it. This also defines inertial frames.
2. Second law: the net force equals the rate of change of linear momentum, \( \vec F = d\vec p/dt \). For constant mass, \( \vec F = m\vec a \). The law holds along each axis separately.
3. Third law: forces occur in equal and opposite pairs acting on two different bodies. The pair acts for the same time and never cancel because they act on different bodies.
4. Units: 1 N = \( 10^5 \) dyne. Weight is \( mg \) and the kg-weight is 9.8 N.

Momentum, Impulse and Variable Mass

Free Body Diagrams and Standard Situations

1. Pick one body, draw only the forces acting on it, choose axes along the acceleration, and write \( \sum F = ma \) along each axis.
2. Lift moving with acceleration a: apparent weight \( N = m(g + a) \) upward acceleration, \( N = m(g - a) \) downward acceleration, zero in free fall.
3. Atwood machine: \( a = \dfrac{(m_1 - m_2)g}{m_1 + m_2} \), tension \( T = \dfrac{2m_1m_2g}{m_1 + m_2} \). A spring balance or pulley holder reads twice this tension if it holds the pulley.
4. Two blocks in contact pushed by F: \( a = F/(m_1+m_2) \), contact force on the second block \( = m_2F/(m_1+m_2) \).
5. Rope of linear density \( \lambda \) pulled by F: \( a = F/M \) and tension at distance x from the pulled end \( T = F(1 - x/L) \).
6. Constraint relation: the component of velocity along an inextensible rope is the same for every point on it. If the rope makes angle \( \theta \) with the horizontal, the block's horizontal speed is \( v/\cos\theta \).
7. Massless pulley with a force F applied on it: the string tension is \( F/2 \).
8. Spring forces do not change instantly when something else is cut or released, but a string tension can change instantly.

Friction

1. Static friction adjusts itself up to a limit: \( f_s \le \mu_sN \). Kinetic friction is \( f_k = \mu_kN \) and is nearly independent of speed and contact area. Usually \( \mu_s > \mu_k \).
2. Angle of friction \( \lambda \): \( \tan\lambda = \mu \). Angle of repose equals the angle of friction.
3. Net contact force \( = \sqrt{N^2 + f^2} = N\sqrt{1+\mu^2} \) when the body is sliding.
4. Friction on a person who walks or runs is forward, because it is the only horizontal external force. Maximum acceleration due to friction is \( \mu g \).
5. Stopping a sliding body: deceleration \( \mu g \), stopping distance \( v^2/2\mu g \), stopping time \( v/\mu g \).

Friction on an Inclined Plane

Blocks Placed on Each Other

1. The upper block is accelerated by friction alone, so its maximum acceleration is \( \mu g \).
2. If the force acts on the lower block (smooth ground), both move together while \( F \le \mu(m_1+m_2)g \).
3. If the ground is rough with coefficient \( \mu_2 \), the force that makes the upper block just slip is \( F = \mu_2(m_1+m_2)g + (m_1+m_2)\mu_1g \).
4. Hanging mass on a table: the block on the table starts to move when \( m_Bg > \mu m_Ag \).
5. Rope on a table with a hanging part: \( l_{hang} = \dfrac{\mu l}{1+\mu} \) at the verge of sliding.

Pseudo Force and Frames

1. In a frame with acceleration \( \vec a_0 \) add a pseudo force \( -m\vec a_0 \) on every body and then use Newton's laws as usual.
2. Force, mass and acceleration are the same in all inertial frames, but velocity, displacement and kinetic energy are not.
3. In a frame of a wedge moving horizontally with acceleration \( a_0 \), resolve gravity and the pseudo force along and perpendicular to the incline.

Springs and Strings

1. Spring force \( F = kx \). Springs in series carry the same force, and \( 1/k_{eq} = 1/k_1 + 1/k_2 \). Springs in parallel share the force and \( k_{eq} = k_1 + k_2 \).
2. A light spring or string has the same tension at both ends. A spring balance reads the tension in it.
3. When a block is hanging on springs and one spring is cut, find the original forces first and then remove the cut spring's force.

Solved Examples

1. A 6000 kg lift accelerates upward at 5 m s\( ^{-2} \). \( T = m(g+a) = 6000\times15 = 90000 \) N.
2. A 2 kg block is stopped by friction in 10 s from 6 m s\( ^{-1} \). \( \mu g = 0.6 \) so \( \mu = 0.06 \).
3. Blocks of 3 kg and 2 kg touch each other and a force of 5 N pushes the first. If a 2 N friction acts on the second, \( a = (5-2)/5 = 0.6 \) m s\( ^{-2} \).
4. Force 25 N on a 10 m rope of 0.5 kg/m: \( a = 5 \) m s\( ^{-2} \), tension 6 m from the pulled end \( = 4\times0.5\times5 = 10 \) N.