I posted this here once before but both answers were wrong. Example 8.5 Kinetic
ID: 1370031 • Letter: I
Question
I posted this here once before but both answers were wrong.
Example 8.5 Kinetic Energy in a Perfectly Inelastic Collision Problem We claimed that the maximum amount of kinetic energy was transformed to other forms in a perfectly inelastic collision. Prove this statement mathematically for a one-dimensional two-particle collision Solution Conceptualize We will assume that the maximum kinetic energy is transformed and prove that the collision must be perfectly inelastic Categorize We categorize the system of two particles as an isolated system. We also categorize the collision as one-dimensional Analyze Find an expression for the ratio of the final kinetic energy after the collision to the initial kinetic energy m1V1 2 2 2 2 2 The maximum amount of energy transformed to other forms corresponds to the minimum value of f. For fixed initial conditions, imagine that the final velocities v1f and v2f are variables. Minimize the fraction f by taking the derivative of f with respect to vif and setting the result equal to zero: dv2f dv1f =0 2 2 dv2f (1) m1v1f + m2v2f_ = 0 From the conservation of momentum condition, we can evaluate the derivative in (1). Differentiate the conservation of momentum condition with respect to V1f. (m1V1f m2V2f) dv1f dv1fExplanation / Answer
(a)
m1 = 3.9 Kg
V1 = 5.10 m/s
m2 = 5.3 kG
v2 = 0
Let the final velocity of both the paticles after collision = Vf
Initial Momentum = m1*v1 + m2*v2
Final Momentum = (m1 + m2) * Vf
Using Momentum Conservation -
Initial Momentum = Final Momentum
m1*v1 + m2*v2 = (m1 + m2) * Vf
3.9 * 5.10 + 0 = (3.9 + 5.3) * Vf
Vf = (3.9 * 5.10 )/(3.9 + 5.3)
Vf = 2.16 m/s
Initial Kinetic Energy = 0.5*m1*v1^2
Final Kinetic Energy = 0.5*(m1 + m2)* Vf^2
Loss of Kinetic Energy = 0.5*m1*v1^2 - 0.5*(m1 + m2)* Vf^2
Loss of Kinetic Energy = 0.5*3.9*5.1^2 - 0.5*(3.9+5.3) * 2.16^2
Loss of Kinetic Energy = 29.25 J
(b)
m1 = 5.3 Kg
V1 = 5.10 m/s
m2 = 3.9 Kg
v2 = 0
Let the final velocity of both the paticles after collision = Vf
Initial Momentum = m1*v1 + m2*v2
Final Momentum = (m1 + m2) * Vf
Using Momentum Conservation -
Initial Momentum = Final Momentum
m1*v1 + m2*v2 = (m1 + m2) * Vf
5.3 * 5.10 + 0 = (3.9 + 5.3) * Vf
Vf = (5.3 * 5.10 )/(3.9 + 5.3)
Vf = 2.94 m/s
Initial Kinetic Energy = 0.5*m1*v1^2
Final Kinetic Energy = 0.5*(m1 + m2)* Vf^2
Loss of Kinetic Energy = 0.5*m1*v1^2 - 0.5*(m1 + m2)* Vf^2
Loss of Kinetic Energy = 0.5*5.3*5.1^2 - 0.5*(3.9+5.3) * 2.94^2
Loss of Kinetic Energy = 29.25 J