Question-136930

Question Number 136930 by I want to learn more last updated on 27/Mar/21

Commented by I want to learn more last updated on 27/Mar/21

Commented by Olaf last updated on 28/Mar/21

(i) W = (1/2)Kx^2 K = ((2W)/x^2 ) = ((2×10)/(0,05^2 )) = 8000 Nm^(−1) (ii) F = Kx = 8000×0,05 = 400 N (iii) F = ma ⇒ a = (F/m) = ((400)/2) = 200 ms^(−2) (iv) V_P = at = 200×0,25 = 50 ms^(−1) M_P V_P + M_Q V_Q = (M_P +M_Q )V 2×50+4×0 = (2+4)V V = 16,7 ms^(−1)

$$\left({i}\right)\:{W}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{Kx}^{\mathrm{2}} \\ $$$${K}\:=\:\frac{\mathrm{2}{W}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}×\mathrm{10}}{\mathrm{0},\mathrm{05}^{\mathrm{2}} }\:=\:\mathrm{8000}\:\mathrm{N}{m}^{−\mathrm{1}} \\ $$$$\left({ii}\right)\:{F}\:=\:{Kx}\:=\:\mathrm{8000}×\mathrm{0},\mathrm{05}\:=\:\mathrm{400}\:\mathrm{N} \\ $$$$\left({iii}\right)\:{F}\:=\:{ma}\:\Rightarrow\:{a}\:=\:\frac{{F}}{{m}}\:=\:\frac{\mathrm{400}}{\mathrm{2}}\:=\:\mathrm{200}\:{ms}^{−\mathrm{2}} \\ $$$$\left({iv}\right)\:{V}_{{P}} \:=\:{at}\:=\:\mathrm{200}×\mathrm{0},\mathrm{25}\:=\:\mathrm{50}\:{ms}^{−\mathrm{1}} \\ $$$${M}_{{P}} {V}_{{P}} \:+\:{M}_{{Q}} {V}_{{Q}} \:=\:\left({M}_{{P}} +{M}_{{Q}} \right){V} \\ $$$$\mathrm{2}×\mathrm{50}+\mathrm{4}×\mathrm{0}\:=\:\left(\mathrm{2}+\mathrm{4}\right){V} \\ $$$${V}\:=\:\mathrm{16},\mathrm{7}\:{ms}^{−\mathrm{1}} \\ $$

Commented by mr W last updated on 28/Mar/21

all is correct except (iv), since the block P doesn′t move with constant acceleration after release. period T=2π(√(m/k))=2π(√(2/(8000)))≈0.1 s for the block P to return back to its initial position it needs (T/4)≈0.025 s. at this position the spring is completely relaxed and losses connection with the block P which then moves with a constant velocity v_P : v_P =(√((2W)/m))=(√((2×10)/2))=(√(10)) m/s also at time t=0.25 s, the block P has this velocity till it collides with block Q. the velocity after collision is v_(P+Q) =((M_P v_P )/(M_P +M_Q ))=((2×(√(10)))/(2+4))=((√(10))/3)≈1.05 m/s

$${all}\:{is}\:{correct}\:{except}\:\left({iv}\right),\:{since}\:{the} \\ $$$${block}\:{P}\:{doesn}'{t}\:{move}\:{with}\:{constant} \\ $$$${acceleration}\:{after}\:{release}. \\ $$$${period}\:{T}=\mathrm{2}\pi\sqrt{\frac{{m}}{{k}}}=\mathrm{2}\pi\sqrt{\frac{\mathrm{2}}{\mathrm{8000}}}\approx\mathrm{0}.\mathrm{1}\:{s} \\ $$$${for}\:{the}\:{block}\:{P}\:{to}\:{return}\:{back}\:{to}\:{its} \\ $$$${initial}\:{position}\:{it}\:{needs}\:\frac{{T}}{\mathrm{4}}\approx\mathrm{0}.\mathrm{025}\:{s}. \\ $$$${at}\:{this}\:{position}\:{the}\:{spring}\:{is}\: \\ $$$${completely}\:{relaxed}\:{and}\:{losses}\: \\ $$$${connection}\:{with}\:{the}\:{block}\:{P}\:{which}\:{then} \\ $$$${moves}\:{with}\:{a}\:{constant}\:{velocity}\:{v}_{{P}} : \\ $$$${v}_{{P}} =\sqrt{\frac{\mathrm{2}{W}}{{m}}}=\sqrt{\frac{\mathrm{2}×\mathrm{10}}{\mathrm{2}}}=\sqrt{\mathrm{10}}\:{m}/{s} \\ $$$${also}\:{at}\:{time}\:{t}=\mathrm{0}.\mathrm{25}\:{s},\:{the}\:{block}\:{P}\:{has} \\ $$$${this}\:{velocity}\:{till}\:{it}\:{collides}\:{with}\:{block} \\ $$$${Q}.\:{the}\:{velocity}\:{after}\:{collision}\:{is} \\ $$$${v}_{{P}+{Q}} =\frac{{M}_{{P}} {v}_{{P}} }{{M}_{{P}} +{M}_{{Q}} }=\frac{\mathrm{2}×\sqrt{\mathrm{10}}}{\mathrm{2}+\mathrm{4}}=\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}\approx\mathrm{1}.\mathrm{05}\:{m}/{s} \\ $$

Commented by I want to learn more last updated on 28/Mar/21