A block of mass m=34 kg has a speed V and is behind a block of mass M = 81-kg that is moving with a speed of 0.50 m/s (both blocks are moving in the same directions). The surface is frictionless and the blocks collide and couple. After the collision, the blocks have a common speed of 0.90 m/s. What is the magnitude of the impulse on the 34-kg block due to the collision? a.32 N. s

b 14 N. s

c 57 N. s

d 73 N. s

e 41 N. s

A., d., and f i think

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Distance covered = speed x timethe distance traveled going against the stream = distance traveled with the stream.therefore it is the speed and the time that are the variables.as he travels the same distance both ways, speed x time is the same for each way.going upstream his speed is 4 kph; going downstream his speed is 6 kph.therefore we have two simultaneous equations: time upstream x 4 = time downstream x 6time upstream + time downstream = 60 minutestherefore we have the equation: time upstream = 6/10 hour, time downstream therefore = 4/10 hour.distance traveled at 4km/hr in 6/10 hour = 24/10 = 2.4km.check: distance traveled at 6km/hr in 4/10 hour = 24/10 = 2.4km.