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nikklg [1K]
3 years ago
11

Two gliders collide on an air track moving in from opposite directions, and they bounce back. The masses of the gliders are 0.20

kg and 0.30 kg. The magnitudes of their initial velocities are 0.50 m/s and 0.40 m/s, respectively. After the collision, the magnitudes of their velocities are 0.40 m/s and 0.20 m/s, respectively. Assume the first glider is initially heading in the positive direction.The total momentum of the gliders before collision is closest to:
Physics
1 answer:
timurjin [86]3 years ago
3 0

Answer:

so initial momentum is 0.22kgm/s

Explanation:

m1=0.20kg

m2=0.30kg

initial velocity of m1=u1=0.50m/s

initial velocity of m2=u2=0.40m/s

total momentum of the system before collision

Pi=m1u1+m2u2

Pi=0.20kg×0.50m/s+0.30kg×0.40m/s

Pi=0.1kgm/s+0.12kgm/s

Pi=0.22kgm/s

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8 0
3 years ago
The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p
NISA [10]

Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

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b) velocity = distance * time

Let the velocity of the swimmer be v_{s} = 1.5 m/s

The separation of the two sides of the river, d = 80 m

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t = 80/1.5

t = 53.33 s

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S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

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cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

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the horizontal speed,

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The downstream horizontal distance of the swimmer, x = v_{x} t

x = 1.6 * 66.67

x = 106.67 m

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