Item 3 Part A A 50.0 g marble moving at 2.10 m/s strikes a 27.0 g marble at rest

Question

Item 3 Part A A 50.0 g marble moving at 2.10 m/s strikes a 27.0 g marble at rest. Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line. What is the speed of 50.0 g marble immediately after the collision? Express your answer with the appropriate units. (viz)50.0g = 1 Value Units Submit My Answers Give Up Part B What is the speed of 27.0 g marble immediately after the collision? Express your answer with the appropriate units. (viz)27.0g = 1 Value Units Submit My Answers Give Up

Explanation / Answer

Let the speed after collision be v1 and v2

by the conservation of momentum we have

m1u1 + m2u2 = m1v1 + m2v2

0.05*2.1 + 0 = 0.05*v1 + 0.027v2

5v1 + 2.7v2 = 10.5 -------(1)

velocity of approach = velocity of recess

v2 - v1 = u1 - u2

v2 - v1 = 2.1 - 0

v2 - v1 = 2.1 ---------(2)

Multiply the (2) by 5 and adding we have

v2 = 2.73 m/s

v1 = v2 - 2.1 = 2.73 - 2.1 = 0.63 m/s

a) 0.63 m/s

b) 2.73 m/s

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