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Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 16.6 kg and an initial velocity of nu Overscript bar EndScripts Subscript 0A = 8.92 m/s, due east. Object B, however, has a mass of mB = 29.2 kg and an initial velocity of nu Overscript bar EndScripts Subscript 0B = 4.44 m/s, due north. Find the (a) magnitude and (b) direction of the total momentum of the twoobject system after the collision.
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2 answers

the vector sum of the initial momentums is equal to the final vector momenmtum
MaVaE+MbVbN=(Ma+Mb)V
well, let Ma' be Ma/(Ma+Mb)
and Mb' be Mb/(Ma+Mb)
V=Ma'*VaE+Mb'*VbN
magnitude..
V^2=(Ma'Va)^2+(Mb'*Vb)^2
direction (degrees N of E)
Theta=arctan((Mb'*Vb)/(Ma'*Va)) 👍
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answered by Anonymous 
M1*V1 + M2*V2 = M1*V + M2*V
16.6*8.92 + 29.2*4.44i = 16.6V + 29.2V
148.1 + 129.6i = 45.8V
196.8[41.2o] = 45.8V
V = 4.3 m/s[41.2o]
a. Momentum = 196.8 kgm/s
b. Direction = 41.2o N. of E. 👍
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answered by Henry
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