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
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.
1 answer
0 answers