Block A in has mass 1.15 kg , and block B has mass 2.70 kg . The blocks are forc

Question

Block A in has mass 1.15 kg , and block Bhas mass 2.70 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 1.25 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

part A:

apply the law of conservation of linear momentum

m1u1 + m2u2 = m1v1 + mv2

here m1u1 + m2u2 = 0

s0 m2v2 = m1v1

v1 = (2.7 * 1.25)/(1.15)

v1 = 2.93 m/s

----------------------------

part B :

PE stored = KE of block 1 + KE of block 2

PE stored = (0.5 * 1.15 * 2.92^3) + (0.5* 2.7 * 2.93^2)

PE stored = 26 J

-------------------------------------

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects