Question

An object of mass 4 Kg is moving towards
the east at a velocity of 6 meters per
second) It collides and sticks to a 6 kg
object moving with a velocity of 5 meters
per second in the same direction. How
much kinetic energy was lost in the
collision​

Answer 1

Answer:

The answer for your question is given in the  photo

Answer 2

Answer:

The kinetic energy equals to 1.2 Joule, was lost in the collision.

Explanation:

Given, the mass of the first object, m₁ = 4Kg

The velocity of first object, v₁ = 6m/s

The mass of the second object, m₂ = 6Kg

The velocity of the second object, v₂ = 5m/s

In the collision, momentum is conserved and a vector quantity. The final velocity 'v':-

$m_1v_1+m_2v_2=mv$

$(4)(6)+(6)(5)=(4+6)v$

$10v=24+30$

$v=5.4m/s$

The change in kinetic energy = Initial K.E. - Final K.E.

Change in K.E. $=\frac{1}{2}(10)(5.4)^2-[(\frac{1}{2}(4)(6)^2)+(\frac{1}{2}(6)(5)^2)]$

Change in K.E. = 145.8 - 72 - 75

Change in K.E. = - 1.2 J

Therefore, the loss in the kinetic energy will be equal to 1.2 Joule.

To learn more about "Law of conservation of momentum"

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