Block A in (Figure 1i has mass 0.750 kg . and block B has mass 3.20 kg . The blo

Question

Block A in (Figure 1i has mass 0.750 kg . and block B has mass 3.20 kg . The blocks are forced together, compressing a spring S between them: then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.50 m/s .What is the final speed of block A? Express your answer in meters per second to three significant figures. How much potential energy was stored in the compressed spring? Express your answer in joules to three significant figures.

Explanation / Answer

ma = 0.75 Kg
mb = 3.2 Kg

vaf = ?
Vbf = 1.5 m/s

(a)
Using Momentum , Conservation =
Initial Momentum = 0
Final Momentum = ma*va + mb*vb

ma*va + mb*vb = 0
0.75 * va = - 3.2 * 1.5
va = - 6.4 m/s (-ve sign indicates it moves in the opoosite direction)

Final Speed of block A = 6.4 m/s

(b)
Final Kinetic Energy of the system,

= 0.5 * ma*va^2 + 0.5 * mb*vb^2
= 0.5 * 0.75*6.4^2 + 0.5*3.2*1.5^2
= 18.96 J

This is equivalent to the Initial Spring Potential Energy

Potential Energy stored in the compressed spring = 18.96 J

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