inelastic collisionkinetic energy loss

Inelastic Collision

  • Inelastic
    Bodies collide, bounce off, eath other move off with seperate velocities.
    Ex: car collison and unstuck
  • Perfectly Inelastic
    Bodies collide and stick together, and have the same final velocity.
    Ex: asteroid hitting earch, bug hitting car windshield

In both case, only momentum is conserved, the kinetic energy is not conserved.

P⃗i=P⃗fm1v1i+m2v2i=m1v1f+m2v2fKE1=12(m1v1i2+m2v2i2)KE2=12(m1v1f2+m2v2f2)ΔKE=KE2−KE1≠0\begin{aligned} \vec P_i &= \vec P_f\\ m_1v_{1i} + m_2v_{2i} &= m_1v_{1f} + m_2v_{2f}\\ KE_1 &= \frac{1}{2}(m_1v_{1i}^2+m_2v_{2i}^2)\\ KE_2 &= \frac{1}{2}(m_1v_{1f}^2+m_2v_{2f}^2)\\ \Delta KE &= KE_2 - KE_1 \ne 0 \end{aligned}

Exercises

(8.21) On a frictionless, horizontal air table, puck A (with mass 0.250 kg) is moving toward puck B (with mass 0.355 kg), that is initially at rest. After the collision, puck A has a velocity of 0.119 m/s to the left, and puck B has velocity 0.650 m/s to the right.
(a) What was the speed of puck A before the collision?
Graph

Solution

Identify: Apply conservation of momentum to the system of the two pucks.
Set up: Let +x+x be to the right.
Execute:

Pi=Pfmavai=mavaf+mbvbfma=0.25kg,vaf=−0.119m/s,mb=0.355kg,vbf=0.650m/svai=mavaf+mbvbfma=−0.119×0.25+0.355×0.6500.25=0.804m/s\begin{aligned} P_i &= P_f\\ m_a v_{ai} &= m_a v_{af} + m_b v_{bf}\\ m_a &= 0.25kg, v_{af} = -0.119 m/s, m_b = 0.355kg, v_{bf} = 0.650 m/s\\ v_{ai} &= \frac{m_a v_{af} + m_b v_{bf}}{m_a}\\ &= \frac{-0.119 \times 0.25 + 0.355\times 0.650}{0.25}\\ &= 0.804m/s \end{aligned}

(b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

Solution

K1=12(mavai2+mbvbi2)=12(0.250×0.8042+0)=8.08×10−2JK2=12(mavaf2+mbvbf2)=12(0.250×(−0.119)2+0.355×0.6502)=7.68×10−2JΔK=K2−K1=−4.04×10−3J\begin{aligned} K_1 &= \frac{1}{2}(m_a v_{ai}^2 + m_b v_{bi}^2)\\ &= \frac{1}{2}(0.250 \times 0.804^2 + 0)\\ &= 8.08 \times 10^{-2} J\\ K_2 &= \frac{1}{2}(m_a v_{af}^2 + m_b v_{bf}^2)\\ &= \frac{1}{2}(0.250 \times (-0.119)^2 + 0.355 \times 0.650^2)\\ &= 7.68 \times 10^{-2} J\\ \Delta K &= K_2 - K_1 = -4.04 \times 10^{-3} J \end{aligned}

(Quiz) A block of mass m=25kgm = 25kg has a speed of vv and is behind of a block M=82kgM=82kg that has a speed of 0.5m/s0.5m/s. THe surface is frictionless. The blocks collide and couple. After the collision, the blocks have a commone speed of 0.9m/s0.9m/s. Find the loss of kinetic energy.
Graph

Solution

1. By Conservation of Momentum

mv+MvM0=(m+M)vfv=(m+M)vf−MvM0m=0.9⋅(25+82))−0.5⋅8225≈2.2m/s\begin{aligned} mv + Mv_{M0} &= (m+M)v_f\\ v &= \frac{(m+M)v_f - Mv_{M0}}{m} \\ &= \frac{0.9 \cdot (25 + 82)) - 0.5 \cdot 82}{25} \approx 2.2m/s \end{aligned}

2. By Conservation of Energy

12mv2+12MvM02=12(m+M)vf2+ElossEloss=12mv2+12MvM02−12(m+M)vf2=12×25×2.22+12×82×0.52−12×(25+82)×0.92=27.42J\begin{aligned} \frac{1}{2}mv^2 + \frac{1}{2}Mv_{M0}^2 &= \frac{1}{2}(m+M)v_f^2 + E_{loss}\\ E_{loss} &= \frac{1}{2}mv^2 + \frac{1}{2}Mv_{M0}^2 - \frac{1}{2}(m+M)v_f^2\\ &= \frac{1}{2} \times 25 \times 2.2^2 + \frac{1}{2} \times 82 \times 0.5^2 - \frac{1}{2} \times (25+82) \times 0.9^2\\ &= 27.42 J \end{aligned}

(Mid Term 2) A block moves at 5.0m/s in the positive x direction and hits an identical block, initial at rest. A small amount of gunpowder had been placed on one of the blockes, The explosion does not harm the blocks but it doubles their total kinectic energy. After the explosion the blocks move along the x axis the incident block has a speed of:

Solution

m1v0=m1v1+m2v212m1v12+12m2v22=2×(12m1v02)m1=m2⇒2v22−10v2−25=0⇒v2=6.83m/s or −1.83m/s\begin{aligned} m_1 v_0 &= m_1 v_1 + m_2 v_2\\ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 &= 2 \times (\frac{1}{2}m_1 v_0^2)\\ m_1 &= m_2\\ \To 2v_2^2 &-10v_2-25 = 0\\ \To v_2 &= 6.83m/s \text{ or } -1.83m/s \end{aligned}

The second block moves along positve axis at v2=6.83m/sv_2 = 6.83m/s; the first block moves along negative x-axis at v1=1.83m/sv_1 = 1.83m/s