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Therefore, the third fragment must fly off toward a general direction of (2) north of east. (b) Apply momentum conservation Po = P. along the xaxis: (2.0 kg)(0) = (0.50 kg)(−2.8 m/s) + (1.3 kg)(0) + (1.2 kg) vx, so vx= 1.17 m/s; along the yaxis: (3.0 kg)(0) = (0.50 kg)(0) + (1.3 kg)(−1.5 m/s) + (1.2 kg) vy, so vy= 1.63 m/s. v= (1.17 m/s)2+ (1.63 m/s)2= 2.0 m/s, θ= tan−1163117..= 54°north of east. 41. (a) Energy conservation gives mgh = 12mv2v= [2(9.80 m/s2)(0.10 m)]1/2= 1.4 m/s(b) The balls stick together, so their speed just after the collision is 0.700 m/s, by momentum conservation. Using energy conservation again gives h = v2/2g= (0.700 m/s)2/[2(9.80 m/s2)] = 0.025 m = 2.5 cm46. m1= 0.010 kg, m2= 3.0 kg, v1o= 400 m/s, v2o= 0, v1= 300 m/s, v2= ? Po= P, m1 v1o+ m2 v2o= m1 v1+ m2 v2. v2= m1 v1o+ m2 v2o−m1 v1m2= (0.010 kg)(400 m/s) + 0 −(0.010 kg)(300 m/s) 3.0 kg= 0.33 m/s. y31x2
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