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.
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
To solve this problem, we need to apply the principles of conservation of energy and conservation of momentum.
(a) To determine the velocity of each block just before the collision, we can use the principle of conservation of energy. At the initial position, the only energy present is gravitational potential energy. Therefore, we can equate the initial potential energy to the kinetic energy just before the collision.
Potential Energy (m1) + Potential Energy (m2) = Kinetic Energy (m1) + Kinetic Energy (m2)
m1 * g * y + m2 * g * y = 1/2 * m1 * v1^2 + 1/2 * m2 * v2^2
Here, g is the acceleration due to gravity, and y is the height of release.
In this case, m1 = 2.00 kg, m2 = 4.50 kg, y = 5.80 m, and g = 9.8 m/s^2.
By solving this equation, you can find the velocities of both blocks just before the collision, denoted as v1 and v2.
(b) To determine the velocities of each block immediately after the collision, you can use the principle of conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Here, v1' and v2' are the velocities of the blocks immediately after the collision.
Solve this equation to find v1' and v2'.
(c) To determine the maximum heights to which m1 and m2 rise after the collision, you can consider the conservation of mechanical energy.
Since the collision is elastic and occurs on a frictionless track, mechanical energy is conserved throughout the system.
To find the maximum heights, we can use the following equation:
Potential Energy (m1) + Potential Energy (m2) = Kinetic Energy (m1) + Kinetic Energy (m2)
m1 * g * h1 + m2 * g * h2 = 1/2 * m1 * v1'^2 + 1/2 * m2 * v2'^2
Here, h1 and h2 are the maximum heights to which m1 and m2 rise after the collision, respectively.
Solve this equation to find h1 and h2.
Make sure to correctly substitute the given values and perform the necessary calculations to obtain accurate answers. Double-check your calculations and units to ensure accuracy.