slides to the left with a velocity of 0.600 m/s on the marble sliding to the rig

Question

slides to the left with a velocity of 0.600 m/s on the marble sliding to the right with a velocity of magnitude 0.150 m/s a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a lne. Take right as the positive x direction.) m/s (smaller marble) m/s (larger marble) kg·m/s (smaller marble) kg m/s (larger marbile) J (smaller marble) (larger marble) s fired at which falls verticaly with zero initial speed. You can ignore ar

Explanation / Answer

8. for elatsic collision,

velocity of separation = velocity of approach

v + V = 0.6 + 0.150

V = 0.75 - v

Applying momentum conservation,

(10 x - 0.60) + (35 x 0.150) = 10v - 35V

- 0.75 = 10 v - 35 (0.75 - v)

v = 0.567 m/s

V = 0.183 m/s

(A) 0.567 m/s ....Smaller

- 0.183 m/s ...Larger

(b) for smaller = 0.010 ( 0.567 - (-0.600)) = 0.01167 kg m/s

for larger = 0.035 ( - 0.183 - 0.150) = - 0.01167 kg m/s

(c) smaller = 0.010 (0.567^2 - 0.6^2) / 2 = - 1.93 x 10^-5 J

larger = 0.035 (0.183^2 - 0.150^2)/2 = 1.93 x 10^- 5 J

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