Before collision, Car A and B were initially traveling Westward and Southward, respectively. The velocity after impact of the two entangled cars is 13.682 m/s. Determine the speed of car B before the collision if it had a mass of 2000 kg and car A had a speed of 30 m/s and a mass of 1400 kg.

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A: Given Mass m1 = 2.6 kg Mass m2 = 1.8 kg Speed v1 = 4.5 m/s Speed v2 = 3.3 m/s

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A: Given m1=20.00 kgv1=5.00 msm2=10.00 kgv2=-00 ms Apply principle of conservation of momentum to find…

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A: Given collision is perfectly elastic so no energy losses are there which allows us to use kinetic…

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A: mass of first object (m1) = 2 kg Initial velocity of first object (u1) = 12 m/s mass of second…

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A: SOlution: Given that mA = 0.6 kg mB = 1.8 kg vA =+4 m/s vB =+2 m/s

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A: in this question  we use conservation law of momentum.

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A: Given that:- pA=9.5×104 kg.km/h (East)pB=1.05×105 kg.km/h (North)m=mass of each truck

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A: In all collision moment remains conserved Momentum before collision = momentum after collision

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A: Dear student We can conserve the momentum when there is no external force acting on the system. The…

Q: An SUV with mass 1.80 × 103 kg is travelling eastbound at +15.0 m/s, while a compact car with mass…

A: (a) Use conservation of momentum m1u1+m2u2=m1+m2vv=m1u1+m2u2m1+m2=1.80×103 kg15.0 m/s+9.00×102…

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A: M =1800 kg  m = 900 kg  U = 15 m/s  u = - 15 m/s

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A: In an inelastic collision, the momentum is conserved and not the energy.Using the conservation of…

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A: Dear student During collision, the momentum remains conserved. Also the coefficient of restitution…

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A: m1 = 4.70 kg v1 = 8.40 m/s m2 = 4 kg v2 = -10.4 m/s v1f = -2.2 m/s

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Q: A 2.5 kg block, initially moving due east at 6 m/s, is acted upon by an impulse having magnitude 15…

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A: Answer

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A: The relationship of impulse and momentum is, I=m∆v I=1500(15-2.6) I=18600 kg m/s

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Q: If a 1305 kg train car on a one dimensional track, moving at 4 m/s north, hits and bounces off a…

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A: From conservation of momentum along x-axis,