Momentum is conserved in collisions Thus pinitial = pfinal Bouncy: mv + m2v2 = m
ID: 1351535 • Letter: M
Question
Momentum is conserved in collisions
Thus
pinitial = pfinal
Bouncy: mv + m2v2 = m vf + m2 v2f
Sticky: mv + m2v2 = (m + m2) vf
mcar1 300 gr
mcar2 500 gr
Kinetic energy is ONLY conserved for elastic
collisions (no bending, no sticking, no appreciable noise, etc.)
KEinitial KEfinal
Bouncy: ½m1v12 + ½m2v22 = ½m1v1f2 + ½m2v2f2
Sticky: ½m1v12 + ½m2v22 ½(m1 + m2) vf2
#1
In Part 1, cars impact and bounce, car 2 is
initially at rest while initial vcar1 = 2 m/s.
(a) If final vcar1 = -½ m/s, what is final vcar2?
(b) Calculate initial KE of the
system in part 1.
(c) Calculate final KE of the
system in part 1
#2
In part 2, cars impact and stick, initial vcar1 is
2.00 m/s while car2 is initially at rest.
(a) What is the velocity of the system
immediately following the collision?
(b) Calculate initial KE of the
system in part 2
(c) Calculate final KE of the
system in part 2
Explanation / Answer
1)
Using momentum coservation,
300 x 2 + 500x0 = (300 x -0.50) + 500v2
v2 = 1.5 m/s
b) initial KE = 0.300 x 2^2 /2 + 0 = 0.6 J
c) final KE = 0.300x0.5^2 /2 + 500 x 1.5^2 /2 = 0.6 J
2)
a) Using momentum coservation,
300 x 2 + 500x0 = (300 + 500)v
v = 0.75 m/s
b) initial KE = 0.300 x 2^2 /2 = 0.6 J
c) final KE = (0.3 + 0.5) x 0.75^2 /2 = 0.225 J
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