A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m = 0.4 k g$ is at rest on this surface. An impulse of $1.0 N s$ is applied to the block at time $t = 0$ so that it starts moving along the x-axis with a velocity $v (t) = v_{0} e^{- t / τ}$ , where $v_{0}$ is a constant and $τ = 4 s$ . The displacement of the block, in metres, at $t = τ$ is Take $e^{- 1} = 0.37$ .

Question

Infinity Learn · Accepted Answer

$$V=V0e$$−$$t/$$τ                    here  $$V0=Jm=2.5m/s$$ ⇒$$dxdt=V0e$$−$$t/$$τ⇒∫$$0xdx=V0$$∫$$0$$τe−$$t/$$τdt           Since∫e−$$xdx=e$$−x−$$1$$⇒$$x=V0e-t/$$τ$$-1$$τ⇒$$x=2.5$$ −$$4$$ e−$$1$$−$$e0$$                  ​⇒$$x=2.5$$−$$4$$ $$0.37$$−$$1$$​⇒$$x=6.30$$

Law of conservation of momentum and it's applications

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Law of conservation of momentum and it's applications

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V=V0e−t/τ here V0=Jm=2.5m/s ⇒dxdt=V0e−t/τ⇒∫0xdx=V0∫0τe−t/τdt Since∫e−xdx=e−x−1⇒x=V0e-t/τ-1τ⇒x=2.5 −4 e−1−e0 ​⇒x=2.5−4 0.37−1​⇒x=6.30

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