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Fudgin [204]
3 years ago
15

Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri

ght and collides with a 1.2 kg mass at rest and the two masses bounce off each other elastically. find the final velocities of the two masses after the collision.
Physics
1 answer:
Elanso [62]3 years ago
8 0

First we can say that since there is no external force on this system so momentum is always conserved.

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}

0.48= 0.6*v_{1f} + 1.2*v_{2f}

0.8  = v_{1f} + 2v_{2f}

now by the condition of elastic collision

v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6

v_{2f} = 0.533 m/s

also from above equation we have

v_{1f} = -0.267 m/s

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

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Explanation:

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Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm

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It is possible because the rubber ball has mechanical energy which is equal to potential energy.

<h3>What is mechanical energy?</h3>

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Kinetic energy is the energy possessed by an object due to its motion.

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Potential energy is the energy possessed by an object due to its position.

When kinetic energy (K.E) = 0

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