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laila [671]
3 years ago
9

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo

ving to the left at 1.0 m/s. What is the total kinetic energy after the collision?
Physics
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

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\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

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