Asked by Corey
Two blocks of masses m1 = 2.00 kg and m2 = 4.50 kg are each released from rest at a height of y = 5.80 m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision.
(a) Determine the velocity of each block just before the collision. Let the positive direction point to the right.
(b) Determine the velocity of each block immediately after the collision.
(c) Determine the maximum heights to which m1 and m2 rise after the collision.
I don't know what I'm doing wrong but I keep getting the wrong answer.
(a) Determine the velocity of each block just before the collision. Let the positive direction point to the right.
(b) Determine the velocity of each block immediately after the collision.
(c) Determine the maximum heights to which m1 and m2 rise after the collision.
I don't know what I'm doing wrong but I keep getting the wrong answer.
Answers
Answered by
Elena
Before collision
PE=KE
m•g•h =m•v²/2
v=sqrt(2•g•h)= v₁₀=v₂₀=sqrt(2•9.8•5.8)=10.7 m/s
(b) After collision
v₁= {-2m₂•v₂₀ +(m₁-m₂)•v₁₀}/(m₁+m₂)=
=(m₁-3m₂)•v/(m₁+m₂)=(2-13.5)•10.7/6.5=
= - 18.93 m/s
v₂={2m₁•v₁₀ - (m₂-m₁)•v₂₀}/(m₁+m₂)=
={3m₁ -m₂)•v/(m₁+m₂)=
=(6-4.5) •10.7/6.5=0.95 m/s
(c)
m•v₁²/2 = m•g•h
h₁= v₁²/2•g=18.93²/2•9.8=18.28 m
m•v₂²/2 = m•g•h
h₂= v₂²/2•g=0.95²/2•9.8=0.046 m
PE=KE
m•g•h =m•v²/2
v=sqrt(2•g•h)= v₁₀=v₂₀=sqrt(2•9.8•5.8)=10.7 m/s
(b) After collision
v₁= {-2m₂•v₂₀ +(m₁-m₂)•v₁₀}/(m₁+m₂)=
=(m₁-3m₂)•v/(m₁+m₂)=(2-13.5)•10.7/6.5=
= - 18.93 m/s
v₂={2m₁•v₁₀ - (m₂-m₁)•v₂₀}/(m₁+m₂)=
={3m₁ -m₂)•v/(m₁+m₂)=
=(6-4.5) •10.7/6.5=0.95 m/s
(c)
m•v₁²/2 = m•g•h
h₁= v₁²/2•g=18.93²/2•9.8=18.28 m
m•v₂²/2 = m•g•h
h₂= v₂²/2•g=0.95²/2•9.8=0.046 m
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.