A $$spring-block$$ system is resting on a frictionless floor as shown in the figure. The spring constant is $$2.0N/m$$ and the mass of the block is $$2.0$$ 𝑘g. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass $$1.0$$ 𝑘𝑔 moving with a speed of $$2.0m/s$$ collides elastically with the first block. The collision is such that the $$2.0$$ 𝑘𝑔 block does not hit the wall. The distance, in meters, between the two blocks when the spring returns to its unstretched position for the first time after the collision is .

Question

A $$spring-block$$ system is resting on a frictionless floor as shown in the figure. The spring constant is  $$2.0N/m$$ and the mass of the block is $$2.0$$ 𝑘g. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass $$1.0$$ 𝑘𝑔 moving with a speed of $$2.0m/s$$  collides elastically with the first block. The collision is such that  the $$2.0$$ 𝑘𝑔 block does not hit the wall.  The distance, in meters, between the two blocks when the spring returns to its unstretched position for the first time after the collision is   .

Infinity Learn · Accepted Answer

$$m1=1kg$$, $$u1=2m/s$$, $$m2=2kg$$, $$u2=0m/s$$​$$v1=m1$$−$$m2m1+m2u1+2m2u2m1+m2=1$$−$$21+22+2$$×$$2$$×$$01+2=23$$​$$v2=m2$$−$$m1m1+m2u2+2m1u1m1+m2=2$$−$$12+10+2$$×$$1$$×$$21+2=43T=2$$$$π$$$$mk=2$$$$π$$$$22=2$$π $$sec$$​Block returns to original position in $$T2=$$πsecDistance covered by $$1kg$$ block in π $$sec$$ is $$d1=v1t=23$$π$$=233.14=2.0933m$$∴$$d1=2.09m$$

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