Vi 2 m/s Mi-M2-200g 1. The visual overview for Case lafter the collision is give
ID: 1483821 • Letter: V
Question
Vi 2 m/s Mi-M2-200g 1. The visual overview for Case lafter the collision is given below: a. Is the momentum conserved? What are the total momenta before and after the collision? b. What are the total mechanical energies before the collision? c. If this mechanical energy is conserved, what are the speeds of the gliders after the collision? Vif 2. The visual overview for Case 2 after the collision is given below: a. Is the momentum conserved? What are the total momenta before and after the collision? Is the momentum conserved? b. What are the total mechanical energies before and after the collision? Is the mechanical energy conserved?Explanation / Answer
a) Yes.
momentum before = M1*v1i
= 0.2*2
= 0.4 kg.m/s
momentum after = mmentum before
= 0.4 kg.m/s
b) moechanical energy before = 0.5*M1*v1i^2
= 0.5*0.2*2^2
= 0.4 J
c) as the two bodies have same masses velocities are interchanged.
v1f = 0
v2f = 2 m/s
2)
a) yes.
initial momentum, Pi = 0.4 kg.m/s
final momentum , Pf= 0.4 kg.m/s
b) initial mechaincal energy = Pi^2/(2*M1)
= 0.4^2/(2*0.2)
= 0.4 J
final mechanical energy = Pf^2/(2*(M1+M2))
= 0.4^2/(2*(0.2+0.2))
= 0.2 J
c) Apply conservation of energy
M1*v1i = (M1+M2)*Vf
==> Vf = M1*v1i/(M1+M2)
= 0.2*2/(0.2+0.2)
= 1 m/s
d) It is a persect inelastic collision.
3) Mechanical energy is not conserved.
loss of mechaincal energy = 0.4 - 0.2
= 0.2
this energy is converted to thermal energy.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.