JEE Physics System Of Particles and Rotational Motion Notes Set 02

Motion of the Centre of Mass

Differentiating \( M\vec R_{cm} = \sum m_i\vec r_i \) gives \( M\vec v_{cm} = \sum m_i\vec v_i = \vec p_{total} \) and \( M\vec a_{cm} = \vec F_{ext} \). The centre of mass moves as if all the mass were concentrated there and all external forces acted on it. Internal forces do not affect it. If \( \vec F_{ext} = 0 \), \( \vec v_{cm} \) is constant; if the system starts from rest, the centre of mass stays at rest. This explains why a man walking on a floating plank moves the plank back, and why the pieces of an exploded bomb or shell keep the original path of the centre of mass.

Impulse and Conservation of Momentum

Impulse \( \vec J = \int\vec F\,dt = \Delta\vec p \). If the net external impulse is zero, the total momentum of a system is conserved. The law can be applied in one direction even when it fails in another. Examples: recoil of a gun, \( \vec V = -\dfrac{m}{M}\vec v \), with kinetic energy \( K = p^2/2m \), so the lighter body gets more kinetic energy; explosion of a bomb at rest (the momenta of the fragments add to zero, and the energy released is the sum of their kinetic energies); a bullet embedded in a block, \( V = \dfrac{mv}{M+m} \).

Collisions and the Coefficient of Restitution

A collision is a short interaction with large internal forces. Momentum is conserved in all collisions. The coefficient of restitution is \( e = \dfrac{\text{speed of separation}}{\text{speed of approach}} \) along the line of impact (no unit).
Elastic: e = 1, kinetic energy conserved.
Inelastic: 0 < e < 1, some kinetic energy lost.
Perfectly inelastic: e = 0, bodies move together, maximum loss.
For a head-on collision, solve \( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \) together with \( v_2 - v_1 = e(u_1 - u_2) \).

Head-On Collision Results

Elastic: \( v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}u_1 + \dfrac{2m_2}{m_1 + m_2}u_2 \), \( v_2 = \dfrac{m_2 - m_1}{m_1 + m_2}u_2 + \dfrac{2m_1}{m_1 + m_2}u_1 \).
Equal masses exchange velocities. A heavy body hitting a light one at rest gives it twice its speed. A light body hitting a heavy one at rest rebounds with the same speed.
Loss of kinetic energy: \( \Delta K = \dfrac{1}{2}\dfrac{m_1m_2}{m_1 + m_2}(1 - e^2)(u_1 - u_2)^2 \). For a perfectly inelastic collision with a stationary target, the fractional loss is \( \dfrac{m_2}{m_1 + m_2} \). The maximum compression of a spring between colliding bodies occurs when both have the same velocity.

Oblique Collisions

Resolve velocities along and perpendicular to the line of impact. Momentum is conserved along the line of impact, and the restitution law applies to those components only; components perpendicular to the line of impact are unchanged for smooth bodies. For an elastic oblique collision between equal masses with one initially at rest, the two bodies move off at right angles to each other. A ball hitting a smooth floor keeps its horizontal component and has its vertical component multiplied by e.

Bouncing Ball

A ball dropped from height h with restitution e: speed after the nth bounce \( v_n = e^n\sqrt{2gh} \), height \( h_n = e^{2n}h \), time of the nth flight \( t_n = e^n\sqrt{2h/g} \). Total time before stopping \( = \sqrt{\dfrac{2h}{g}}\dfrac{1+e}{1-e} \); total distance \( = h\dfrac{1+e^2}{1-e^2} \).

Worked Examples

Worked Example 1. A 4 kg body moving at 6 m/s collides head-on with a 2 kg body at rest. Find the final velocities if e = 0.5.
Answer:
\( 4(6) = 4v_1 + 2v_2 \Rightarrow 2v_1 + v_2 = 12 \) and \( v_2 - v_1 = 0.5\times 6 = 3 \).
Solving: \( v_1 = 3 \) m/s and \( v_2 = 6 \) m/s.

Worked Example 2. A 20 g bullet moving at 250 m/s embeds in a 0.98 kg block. Find the common speed and the fraction of kinetic energy lost.
Answer:
\( V = \dfrac{0.02\times 250}{1} = 5 \) m/s.
Fraction lost \( = \dfrac{M}{m + M} = \dfrac{0.98}{1} = 98\% \).

Worked Example 3. A bomb at rest explodes into two parts of masses 2 kg and 3 kg. The lighter part moves at 30 m/s. Find the energy released.
Answer:
The heavier part has speed \( 2\times 30/3 = 20 \) m/s.
\( E = \tfrac{1}{2}(2)(900) + \tfrac{1}{2}(3)(400) = 900 + 600 = 1500 \) J.