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

Question

Coleman Hardin

Physics

29 July, 21:15

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

+2

Answers (1)

Know the Answer?

Chica · Answer 1

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