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A body of mass m₁ = 1kg is released from rest from the top of the first slope 5m high. After going down the first slope, the first body collides with another stationary body of mass m₂ = 2kg on a flat part. To what height will body m and body m₂ rise after the collision, if it is a perfectly elastic collision and there is no friction on the surfaces?

A body of mass m₁ = 1kg is released from rest from the top of the first slope 5m high-example-1

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To determine the height to which bodies m₁ and m₂ will rise after the perfectly elastic collision, we can use the principle of conservation of mechanical energy.

Before the collision, the only form of energy possessed by the system is gravitational potential energy. The potential energy of body m₁ at the top of the first slope is given by:

PE₁ = m₁ * g * h₁

Where:

m₁ = mass of body m₁ = 1 kg

g = acceleration due to gravity = 9.8 m/s²

h₁ = height of the first slope = 5 m

After the collision, the system will consist of both bodies rising to a certain height, and the total mechanical energy of the system will be equal to the initial potential energy. We can express this as:

PE₁ = m₁ * g * h₁ = PE₂ + PE₃

Where:

PE₂ = potential energy of body m₁ after the collision

PE₃ = potential energy of body m₂ after the collision

Since it is a perfectly elastic collision, the kinetic energy of the system is conserved as well. Therefore, we can express the kinetic energy of the system before and after the collision as:

KE₁ = KE₂ + KE₃

Initially, body m₁ is at rest, so its initial kinetic energy is zero:

KE₁ = 0

After the collision, both bodies will have kinetic energy due to their upward motion. The kinetic energy of a body is given by:

KE = (1/2) * m * v²

Where:

m = mass of the body

v = velocity of the body

Since the collision is perfectly elastic, the velocities of bodies m₁ and m₂ after the collision can be determined by using the principle of conservation of momentum:

m₁ * v₁ = -m₂ * v₂

Where:

v₁ = velocity of body m₁ after the collision

v₂ = velocity of body m₂ after the collision

From the conservation of momentum equation, we can solve for v₁ in terms of v₂:

v₁ = (-m₂ / m₁) * v₂

Substituting this expression for v₁ into the equation for kinetic energy:

KE₂ = (1/2) * m₁ * [(-m₂ / m₁) * v₂]²

= (1/2) * (m₂ / m₁)² * m₁ * v₂²

= (1/2) * (m₂ / m₁)² * KE₃

Since the kinetic energy of the system is conserved, we have:

KE₁ = KE₂ + KE₃

0 = KE₂ + KE₃

0 = (1/2) * (m₂ / m₁)² * KE₃ + KE₃

0 = [(1/2) * (m₂ / m₁)² + 1] * KE₃

Since the kinetic energy cannot be zero, the factor [(1/2) * (m₂ / m₁)² + 1] must be zero. Solving for the height to which body m₂ will rise after the collision:

PE₃ = m₂ * g * h₂

Where:

h₂ = height to which body m₂ will rise after the collision

Setting [(1/2) * (m₂ / m₁)² + 1] = 0:

(1/2) * (m₂ / m₁)² + 1 = 0

(m₂ / m₁)² = -2

m₂ / m₁ = ±√(-2)

Since the square root of a negative number is imaginary, there is no real solution for

the height to which body m₂ will rise after the collision. Therefore, the bodies will not rise after the collision.

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