Two blocks of masses m1 = 2kg. and m2 = 4kg. are each released from rest at a height of h = 5m. on a frictionless track, and undergo an elastic head-on collision. a)Determine the velocity of each block just before the collision. 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.

Question

Two blocks of masses m1 = 2kg. and m2 = 4kg. are each released from rest at a height of h = 5m. on a frictionless track, and undergo an elastic head-on collision. a)Determine the velocity of each block just before the collision. 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.

Answer 1

Kutreya

Kutreya
Kutreya
Kutreya
Kutreya

Answer 2

To solve this problem, we can use the principle of conservation of mechanical energy, along with the conservation of momentum.

a) The velocity of each block just before the collision:

Since the blocks are released from rest at a height of 5m, we can use the equation of gravitational potential energy to find their initial velocities:

m1*gh = (1/2)*m1*v1^2
(2kg)*(9.81m/s^2)*(5m) = (1/2)*(2kg)*v1^2
98.1 = v1^2
v1 = √98.1 m/s ≈ 9.905 m/s

m2*gh = (1/2)*m2*v2^2
(4kg)*(9.81m/s^2)*(5m) = (1/2)*(4kg)*v2^2
196.2 = v2^2
v2 = √196.2 m/s ≈ 14.002 m/s

Therefore, the initial velocity of m1 is approximately 9.905 m/s and the initial velocity of m2 is approximately 14.002 m/s.

b) The velocity of each block immediately after the collision:

Since the collision is elastic, both momentum and kinetic energy will be conserved.

Conservation of momentum:
m1*v1 + m2*v2 = m1*v1' + m2*v2'

Substituting the given values:
(2kg)*(9.905 m/s) + (4kg)*(14.002 m/s) = (2kg)*v1' + (4kg)*v2'
19.81 + 56.008 = 2v1' + 4v2'
75.818 ≈ 2v1' + 4v2' (Equation 1)

Conservation of kinetic energy:
(1/2)*m1*(v1^2) + (1/2)*m2*(v2^2) = (1/2)*m1*(v1'^2) + (1/2)*m2*(v2'^2)

Substituting the given values:
(1/2)*(2kg)*(9.905 m/s)^2 + (1/2)*(4kg)*(14.002 m/s)^2 = (1/2)*(2kg)*(v1'^2) + (1/2)*(4kg)*(v2'^2)
97.754 + 392.179 = (2kg)*(v1'^2) + (4kg)*(v2'^2)
489.933 ≈ 2v1'^2 + 4v2'^2 (Equation 2)

We now have two equations (Equation 1 and Equation 2) with two unknowns (v1' and v2'). We can solve this system of equations to find the values of v1' and v2'.

c) The maximum heights to which m1 and m2 rise after the collision:

After the collision, the blocks will separate and rise to different heights. We can use the principle of conservation of mechanical energy to find these heights. The initial mechanical energy is equal to the final mechanical energy.

Initial mechanical energy = Final mechanical energy
(m1 + m2)*g*h = (1/2)*m1*(v1'^2) + (1/2)*m2*(v2'^2) + (m1 + m2)*g*h'

Substituting the given values:
(2kg + 4kg)*(9.81 m/s^2)*5m = (1/2)*(2kg)*(v1'^2) + (1/2)*(4kg)*(v2'^2) + (2kg + 4kg)*(9.81 m/s^2)*h'

98.1 + 196.2 = v1'^2 + 2v2'^2 + 78.45*h'

We can use the values of v1' and v2' obtained from solving the system of equations above to find the maximum heights h' to which m1 and m2 will rise.

Please note that solving the system of equations requires numerical computation.

Answer 3

To solve this problem, we need to apply the principles of conservation of energy and conservation of momentum. Let's go through each part step by step:

a) Determine the velocity of each block just before the collision:
Since both blocks are released from rest at a height of 5m, we can use the principle of conservation of energy to find the velocity just before the collision.
The potential energy at the starting height will be converted into kinetic energy just before the collision.

For block m1:
Potential Energy (PE1) = m1 * g * h
Kinetic Energy (KE1) = (1/2) * m1 * v1^2

Setting PE1 = KE1, we can solve for v1:
m1 * g * h = (1/2) * m1 * v1^2
v1^2 = 2 * g * h
v1 = sqrt(2 * g * h)

For block m2:
Similar to m1, we have:
v2 = sqrt(2 * g * h)

b) Determine the velocity of each block immediately after the collision:
Since it is an elastic collision, both momentum and kinetic energy will be conserved.

Using the principle of conservation of momentum, we have:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
(where v1i and v2i are the initial velocities, and v1f and v2f are the final velocities after the collision)

Using the principle of conservation of kinetic energy, we have:
(1/2) * m1 *v1i^2 + (1/2) * m2 * v2i^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2

Since both blocks are initially at rest, we have:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
v1f = (m1 - m2) / (m1 + m2) * v1i
v2f = (2 * m1) / (m1 + m2) * v1i

c) Determine the maximum heights to which m1 and m2 rise after the collision:
After the collision, the blocks will separate and rise to their respective maximum heights before coming to rest.

For m1:
Using the principle of conservation of energy (assuming no external forces), potential energy at maximum height will be equal to the initial kinetic energy:
(1/2) * m1 * v1f^2 = m1 * g * h1
h1 = v1f^2 / (2 * g)

For m2:
Using the same principle of conservation of energy, we have:
(1/2) * m2 * v2f^2 = m2 * g * h2
h2 = v2f^2 / (2 * g)

Now you have the equations to solve for the velocities of the blocks before and after the collision, as well as the maximum heights they reach after the collision. Substitute the given values into the equations to obtain the numerical answers.

Answer 4

m₁gh= m₁v₁₀²/2 => v₁₀=sqrt(2gh)

m₂gh= m₂v₂₀²/2 => v₂₀ =sqrt(2gh)

m₁v₁₀ - m₂v₂₀= - m₁v₁ + m₂v₂
v₁= {-2m₂v₂₀ +(m₁-m₂)v₁₀}/(m₁+m₂)
v₂={ 2m₁v₁₀ - (m₂-m₁)v₂₀}/(m₁+m₂)

m₁v₁²/2= m₁gh₁ => h₁=v₁²/2g
m₂v₂²/2= m₂gh₂ => h₁=v₁²/2g