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A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a block of mass M =11.5 kg that is at rest. After the collision, the 4.8-kg block reverses direction, and its new speed is Vf=2.5m/s. What is V, the speed of the 11.5-kg block? 5.6 m/s
6.5 m/s
3.7 m/s
4.7 m/s
4.1 m/s

Respuesta :

gilcarusolautaro

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

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