JEE Physics Centre of Mass Notes Set 06

Impulse and Momentum

1. Linear momentum \( \vec p = m\vec v \). Newton's second law is \( \vec F = \frac{d\vec p}{dt} \).
2. Impulse of a force is \( \vec J = \int \vec F\,dt \), the area under the F-t graph. The impulse-momentum theorem: \( \vec J = \Delta\vec p = m(\vec v - \vec u) \).
3. For an impulsive force such as a bat striking a ball, the force is large and acts for a very short time. Other finite forces (weight, spring forces) produce negligible impulse during the contact and are ignored.
4. Average force \( F_{av} = \frac{J}{\Delta t} \). A longer stopping time reduces the force for the same change of momentum (cricket catch, airbag).
5. A ball of mass m hitting a fixed wall at angle \( \theta \) to the normal with speed v and rebounding elastically gets an impulse \( 2mv\cos\theta \) along the normal; the tangential component of the velocity is unchanged for a smooth wall.
6. The force from a stream of fluid on a wall: with density \( \rho \), area A and speed v, the rate of mass arrival is \( \rho Av \), so the force on a wall that stops the stream is \( \rho Av^2 \), and if the stream rebounds with the same speed it is \( 2\rho Av^2 \).

Conservation of Linear Momentum

1. If the net external force on a system is zero in a direction, the total momentum in that direction is conserved: \( \sum m_i\vec v_i = \text{constant} \).
2. It holds in explosions, collisions and recoil. In a gun firing a bullet, the recoil speed of the gun \( V = \frac{mv}{M} \). Bullet and gun have equal and opposite momenta but the kinetic energy is shared in inverse ratio of the masses, \( \frac{K_{bullet}}{K_{gun}} = \frac Mm \).
3. A body at rest that splits into two parts gives parts moving in opposite directions with equal momenta. If it splits into three, their momenta lie in a plane and add to zero.
4. Momentum is conserved during a collision even when the kinetic energy is not, since the internal forces are equal and opposite.

Collisions in One Dimension

1. Line of impact: the common normal at the contact point. In a head-on (direct) collision the velocities lie along it.
2. Coefficient of restitution \( e = \frac{\text{speed of separation}}{\text{speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2} \) along the line of impact. It lies between 0 (perfectly inelastic) and 1 (perfectly elastic).
3. Final velocities for masses m1 and m2 with initial velocities u1 and u2:
\( v_1 = \frac{(m_1 - em_2)u_1 + (1+e)m_2u_2}{m_1+m_2} \) and \( v_2 = \frac{(m_2 - em_1)u_2 + (1+e)m_1u_1}{m_1+m_2} \).
4. Elastic collision (e = 1): kinetic energy is conserved. Equal masses exchange velocities. A heavy body hitting a light body at rest sends it away at nearly twice its own speed. A light body hitting a heavy one at rest bounces back at nearly the same speed.
5. Perfectly inelastic collision (e = 0): the bodies move together with \( v = \frac{m_1u_1 + m_2u_2}{m_1+m_2} \). The loss of kinetic energy is the largest possible and equals \( \frac12\mu(u_1 - u_2)^2 \).
6. Loss of kinetic energy for any e: \( \Delta K = \frac12\mu(1 - e^2)(u_1 - u_2)^2 \).
7. In an elastic collision with a stationary target of equal mass the fraction of energy transferred is 1; for general masses it is \( \frac{4m_1m_2}{(m_1+m_2)^2} \).
8. Ball dropped from height h onto the floor with restitution e: speed after the nth bounce \( e^n\sqrt{2gh} \), height after the nth rebound \( e^{2n}h \), and the total time before it stops is \( \sqrt{\frac{2h}{g}}\cdot\frac{1+e}{1-e} \).

Oblique Collisions

1. Resolve the velocities along the line of impact and perpendicular to it. For smooth bodies the components perpendicular to the line of impact do not change.
2. Along the line of impact use momentum conservation and the restitution equation, as in the direct case.
3. A ball hitting a smooth fixed wall at angle \( \theta \) to the normal: the component along the wall is unchanged and the normal component is multiplied by \( -e \). The angle of rebound satisfies \( \tan\theta' = \frac{\tan\theta}{e} \).
4. In an elastic oblique collision of equal masses, one at rest, the two bodies move off at right angles to each other.
5. Collision with a wall: the wall absorbs momentum, so momentum of the ball alone is not conserved, but the wall (Earth) gains negligible kinetic energy.

Variable Mass and Rocket Propulsion

1. When the mass of a system changes by ejecting or taking up material, apply Newton's law to the main body and the material together. If material is ejected at relative velocity \( \vec u_{rel} \) and no other force acts, \( M\frac{d\vec v}{dt} = \vec u_{rel}\frac{dM}{dt} \) plus external forces.
2. The thrust on a rocket is \( F = u\left|\frac{dm}{dt}\right| \), where u is the exhaust speed relative to the rocket and \( \left|\frac{dm}{dt}\right| \) is the burn rate.
3. Rocket acceleration in a gravitational field: \( a = \frac{u}{m}\left|\frac{dm}{dt}\right| - g \), where m is the instantaneous mass.
4. Rocket equation in free space: \( v = v_0 + u\ln\frac{m_0}{m} \). In a uniform gravitational field for a vertical flight: \( v = u\ln\frac{m_0}{m} - gt \).
5. For a chain or rope falling on a table, the force on the table is the weight of the part already lying plus the momentum change of the arriving part, \( \lambda v^2 \). At the moment a length x has landed from a free fall the reading is \( 3\lambda xg \) for a chain dropped from rest.
6. A conveyor belt receiving material at a rate \( \mu \) kg/s at speed v needs an extra force \( F = \mu v \) to keep a steady speed.

Worked Examples

1. A 0.2 kg ball moving at 10 m/s is hit back along the same line with the same speed. The impulse is \( 0.2\times10 - 0.2\times(-10) = 4 \) N s. If the contact lasts 0.01 s, the average force is 400 N.
2. A 2 kg mass at 6 m/s hits a 4 kg mass at rest with e = 0.5. The final velocities: \( v_1 = \frac{(2 - 2)\times6}{6} = 0 \) and \( v_2 = \frac{1.5\times2\times6}{6} = 3 \) m/s. Kinetic energy falls from 36 J to 18 J.
3. A ball is dropped from 4 m and rebounds with e = 0.5. The speed after the first bounce is half of the impact speed, so the rebound height is \( e^2h = 1 \) m.
4. A bullet of mass 20 g at 500 m/s embeds in a block of mass 980 g at rest on a smooth surface. Common speed \( = \frac{0.02\times500}{1} = 10 \) m/s. Fraction of kinetic energy lost \( = 1 - \frac{m}{M+m} = 98\% \).
5. A rocket of mass 500 kg ejects gas at 2000 m/s relative to it at 5 kg/s. Thrust \( = 2000\times5 = 10000 \) N. Initial acceleration near the surface \( = \frac{10000}{500} - 10 = 10 \) m/s² (g = 10 m/s²).
6. A particle of mass m collides elastically head-on with a stationary particle of mass 3m. The incoming particle rebounds with \( \frac{m - 3m}{4m}u = -\frac u2 \) and the other moves forward at \( \frac{2m}{4m}u = \frac u2 \).