Help with physics?

2018-07-16 2:54 pm

In the figure, block 1 slides along an x axis on a frictionless floor with a speed of 1.11 m/s. When it reaches stationary block 2, the two blocks undergo an elastic collision. Block 1 has mass 0.317 kg and at t = 0 its center is at x = -1.50m. Block 2 has mass 0.475 kg at t = 0 its center is at x = 0m. Where is the center of mass(in m) of the two block system located at t = 4.70s? (Hint: The intent of this problem is for you to find the initial position of the center of mass and the velocity of the center of mass and use these to find the final position of the center of mass. You do not need to use the elastic collision equations, which would make this problem much longer and more complicated.)

回答 (1)

胡雪8°

2018-07-16 4:11 pm

Time at the collision, t = (1.50 m) / (1.11 m/s) = 1.35 s

Total momentum before collision = Total momentum after collision
(0.317 kg) × (1.11 m/s) + (0.475 kg) × (0 m/s) = (0.317 + 0.475 kg) × v
Velocity after collision, v = 0.317 × 1.11 / (0.317 + 0.475) m/s = 0.444 m/s

At t = 4.70, time taken after collision = 4.70 - 1.35 s = 3.35 s
Displacement at (t = 4.70) = (0.444 m/s) × (3.35 s) = 1.49 m

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