 # Power and Collisions

• Power is the rate of doing work.
• \tt P = \frac{w}{t} (or) \tt P = \frac{dw}{dt}
• Power is a scalar quantity.
• Power P = Force · Velocity = \tt \overline{F} \cdot \overline{V}
• Erg/Sec is the CGS unit of power.
• Walt is the MKS unit of power.
• Average power \tt P_{avg} = \frac{w}{t} = \frac{\frac{1}{2} mv^{2}}{t} = \frac{F \cdot V}{2}
• If a particle moves with constant velocity Pavg = Pinst
• Power of machine gun \tt P = \frac{1}{2} \frac{mnv^{2}}{t}
• Conveyor belt is moving with velocity ‘v’ Gravel is poured at \tt \frac{dm}{dt} rate \tt P = \left(\frac{dm}{dt}\right) v^{2}
• Power of a motor to throw water of density ‘ρ’ with velocity ‘v’ from a tube of area of crossection ‘A’ is P = A ρv3.
• Power of a motor to lift water of mass ‘m’ to a height ‘h’ in time ‘t’ and pumps with velocity V.
\tt P = \frac{mgh + \frac{1}{2} mv^{2}}{t}
• Power of car engine P = (R + ma) v {R = Resistance offered}.
• To get ‘n’ times of water force must be increased by n2 times F1 = n2F.
• To get ‘n’ times of water power must be increased by n3 times P1 = n3P.
• When power of engine is constant velocity \tt V \propto t^{\frac{1}{2}}.
• When power of engine is constant displacement \tt S \propto t^{\frac{3}{2}}.
• 1 watt = \tt \frac{1 Joule}{1 sec}
• 1 Horse power = 746 watt.
• 1 Kilo watt hour = 3.6 × 106 Joules.
• Efficiency \tt \eta = \frac{Power \ output}{Power \ input}
• Gravitational potential energy \tt U = \frac{mgh}{\left(1 + h/R\right)}
• Electrical potential energy \tt U = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}q_{2}}{r}
• Block is pulled by a constant power over a surface with coefficient of friction (μ). The velocity V = \frac{P}{\mu mg}.
• Power can also be represented as P = Fv cos θ.
• A collision is a redistribution of momenta ignoring the effect of other forces with or without physical interaction.
• In a collision there need not be a physical contact between two entities.
• A elastic collision is that in which both momentum and kinetic energy are conserved.
• Collision between sub atomic particles is the example for elastic collision.
• An inelastic collision is that in which only momentum is conserved and not the kinetic energy.
• Collision between automobiles is an example for inelastic collision.
• A collision which takes place along a straight line is called Head on (or) one dimensional collision.
• A oblique collision is that in which before and after the collision are not along the same line.
• Coefficient of restitution (e) is defined as the ratio of relative velocity of separation after collision to relative velocity of approach before collision.
• Coefficient of restitution (e) is a pure number.
• Coefficient of restitution (e) has no units and dimensions.
• Coefficient of restitution (e) is termed as degree of elasticity.
• Practically "e" lies between 0 and 1.
• For perfectly elastic collision e = 1.
• For perfectly inelastic collision e = 0.
• For semi elastic collision 0 < e < 1.
• The value of ‘e’ depends upon nature of colliding bodies.
• When a body is dropped from ‘h1’ (or) bounces to height ‘h2\tt e = \sqrt{\frac{h_{2}}{h_{1}}}
• Height after nth bounce hn = e2n h0.
• If a body incidents with velocity "v" and bounces back with velocity v'. \tt e = \frac{V^{'}}{V}
• If a body takes a time ‘t’ to travel and ‘t'’ to bounce back \tt e = \frac{t^{'}}{t}.
• Time taken for nth bounce tn = en · t.
• Velocity after nth bounce vn = en · v.
• If a ball takes ‘n’ bounces then total distance travelled \tt d = H \left(\frac{1+e^{2}}{1-e^{2}}\right).
• If a ball takes ‘n’ bounces then total time takes to travel \tt T = \sqrt{\frac{2H}{g}} \left(\frac{1+e}{1-e}\right).
• Average speed for ‘n’ bounces = \tt \frac{H \left(\frac{1+e^{2}}{1-e^{2}}\right)}{\sqrt{\frac{2+1}{g}}\left(\frac{1+e}{1-e}\right)}.
• Average velocity for ‘n’ bounces = \tt \frac{H}{\sqrt{\frac{2H}{g}} \left(\frac{1+e}{1-e}\right)}.
• Final velocity of first body after one dimensional elastic collision \tt V_{1} = \left(\frac{m_{1}-m_{2}}{m_{1} + m_{2}}\right) u_{1} + \frac{2 m_{2}u_{2}}{m_{1} + m_{2}}.
• For second body \tt V_{2} = \frac{m_{2}-m_{1}}{m_{1} + m_{2}} \left(u_{2}\right) + \frac{2 m_{1}u_{1}}{m_{1} + m_{2}}.
• Loss in kinetic energy = \tt \frac{1}{2} \frac{m_{1}m_{2}}{m_{1} + m_{2}} \left(u_{1} - u_{2}\right)^{2} = 0 in elastic collision.
• Common velocity of compound body after perfectly inelastic collision is \tt V = \frac{m_{1}u_{1} + m_{2}u_{2}}{m_{1} + m_{2}}.
• Loss in KE in elastic collision = \tt \frac{1}{2} \frac{m_{1}m_{2}}{m_{1} + m_{2}} \left(u_{1} - u_{2}\right)^{2}.
• Final velocity for semi elastic collision. \tt V_{1} = \left(\frac{m_{1} - em_{2}}{m_{1} + m_{2}}\right) u_{1} + \left(\frac{1+e}{m_{1} + m_{2}}\right) m_{2}u_{2}.
• \tt V_{2} = \left(\frac{m_{e} - em_{1}}{m_{1} + m_{2}}\right) u_{2} + \left(\frac{1+e}{m_{1} + m_{2}}\right) m_{1}u_{1}
• Loss in KE in semi elastic collision. \tt \frac{1}{2} \left(\frac{m_{1} m_{2}}{m_{1} + m_{2}}\right) \left(1-e^{2}\right) \left(u_{1} - u_{2}\right)^{2}.
• In perfectly inelastic collision the ratio of loss of energy of the system and its initial energy if u2 = 0 \tt \frac{\Delta KE}{Ei}= \frac{m_{2}}{m_{1} + m_{2}}.
• In perfectly inelastic collision the ratio of final energy to initial energy \tt \frac{Ef}{Ei} = \frac{m_{1}}{m_{1} + m_{2}}.
• In elastic collision ‘m2’ is at rest KE of m1 before collision is E then kinetic energy of m1 after collision \tt K_{1} = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right)^{2} E.
• In elastic collision ‘m2’ is at rest and KE of m1 before collision is E then kinetic energy of m2 after collision \tt K_{2} = \frac{4m_{1}m_{2}}{\left(m_{1} + m_{2}\right)^{2}} E.
• In the above case fraction of KE retained by "m1" is \tt \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right)^{2}.
• Fraction of KE retained by m2= \tt \frac{4m_1m_2}{\left(m_{1} + m_{2}\right)^{2}}.
• If target particle is very light v1 = u1 & v2 = 2u1 (u2 = 0).
• Velocity of bullet from Ballistic pendulum \tt V_{bullet} = \frac{\mu + m}{m} \sqrt{2gh}.
• For oblique impact of smooth spheres.
v1 sin θ1 = u1 sin α1
v2 sin θ2 = u2 sin α2
• m1v1 cos θ1 + m2v2 cos θ2 = m1u1 cos α1 + m2u2 cos α2
• \tt e = \frac{\left(v_{2} \cos \theta_{2} - v_{1} \cos \theta_{1}\right)}{\left(u_{1} \cos \alpha_{1} - u_{2} \cos \alpha_{2}\right)}.
• \tt V_{1} \cos \theta_{1} = \frac{\left(m_{1} - em_{2}\right) u_{1} \cos \alpha_{1} + \left(1 + e\right) m_{2} u_{2} \cos \alpha}{\left(m_{1} + m_{2}\right)}.
• \tt V_{2} \cos \theta_{2} = \frac{\left(1 + e\right) m_{1}u_{1} \cos \alpha_{1} + \left(m_{2} - em_{1}\right) u_{2} \cos \alpha_{2}}{\left(m_{1} + m_{2}\right)}.
• \tt \Delta KE=\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} \left(1 - e^{2}\right) \left(u_{1} \cos \alpha_{1} - u_{2} \cos \alpha_{2}\right)^{2}.
• Collision between blocks with spring \tt E = \frac{1}{2} m_{1}{u_{1}^{2}} + = \frac{1}{2} m_{2}{u_{2}^{2}} + \frac{1}{2} Kx^{2}
• Change in momentum along Normal to spheres = m(v cos α + u cos θ) = m (1 + e) u cos θ.
• Momentum principle in collision.\tt P^{2} = P_{1}^{2} + P_{2}^{2} + 2P_{1}P_{2} \cos \theta.
• Two particles m1 & m2 attached to a string on either ends or an impulse J is given as shown. Then final velocity of m2
\tt V_{2} = \frac{J \cos \theta}{m_{1} + m_{2}}. • Final velocity of m1 \tt V_{1} = \frac{J \sin \theta}{m_{1}}.

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1. Average power (Pav.) = \tt \frac{\Delta W}{\Delta t}=\frac{W}{t}

2. Instantaneous power (Pinst.) = \tt \frac{dW}{dt}=\frac{\overrightarrow{F}.d\overrightarrow{s}}{dt}
Pinst = \overrightarrow{F}.\overrightarrow{v}

3. Inelastic one Dimensional collision
Velocities after collision v_{1}=\frac{\left(m_{1}-m_{2}\right)u_{1}+2 m_{2}u_{2}}{\left(m_{1}+m_{2}\right)}\ and\ v_{2}=\frac{\left(m_{2}-m_{1}\right)u_{2}+2 m_{1}u_{1}}{\left(m_{1}+m_{2}\right)}

4. Loss of kinetic energy \tt \Delta E= \frac{m_{1}m_{2}}{2\left(m_{1}+m_{2}\right)}\left(u_{1}-u_{2}\right)^2\left(1-e^{2}\right)

5. In perfectly Inelastic one Dimensional collision : Velocity of separation after collision = 0
Loss of kinetic energy = \tt \frac{m_{1}m_{2}\left(u_{1}-u_{2}\right)^2}{2\left(m_{1}+m_{2}\right)}

6. If a body is dropped from a height h0 and it strikes the ground with velocity v0 and after inelastic collision it rebounds with velocity v1 and rises to a height h1, then
e=\frac{v_{1}}{v_{0}}=\sqrt{\frac{2gh_{1}}{2gh_{0}}}=\sqrt{\frac{h_{1}}{h_{0}}}

7. If after n collisions with the ground, the body rebounds with a velocity vn and rises to a height hn, then
e^{n}=\frac{v_{n}}{v_{0}}=\sqrt{\frac{h_{n}}{h_{0}}}

8. Two Dimensional or Oblique collision: In horizontal direction, m_{1}u_{1}\cos\alpha_{1}+m_{2}u_{2}\cos\alpha_{2}=m_{1}v_{1}\cos\beta_{1}+m_{2}v_{2}\cos\beta_{2} 9. In vertical direction, m_{1}u_{1}\sin\alpha_{1}-m_{2}u_{2}\sin\alpha_{2}=m_{1}v_{1}\sin\beta_{1}+m_{2}v_{2}\sin\beta_{2} 10. Perfectly Inelastic collision:
When the colliding bodies are moving in the same direction
Loss in kinetic energy \Delta K=\frac{1}{2}\left(\frac{m_{1}m_{2}}{m_{1}+m_{2}}\right)\left(u_{1}-u_{2}\right)^2

11. When the colliding bodies are moving in the opposite direction
Loss in kinetic energy = \frac{1}{2}\ \frac{m_{1}m_{2}}{m_{1}+m_{2}}\ \left(u_{1}-u_{2}\right)^2

12. Velocity of block after collision: v=\frac{mu}{\left(m+M\right)} 13. Velocity of bullet after collision: u=\left[\frac{\left(m+M\right)\sqrt{2gh}}{m}\right]

14. Loss in kinetic energy \Delta K=\frac{1}{2}\ \frac{mM}{m+M}\ u^{2}

15. Angle of string from the vertical \theta=\cos^{-1}\left[1-\frac{1}{2gL}\left(\frac{mu}{m+M}\right)^2\right]