I\'ve been having trouble with #14. Block A, mass 5kg, and block B, mass 1kg, ar

Question

I've been having trouble with #14.

Block A, mass 5kg, and block B, mass 1kg, are sitting on a frictionless surface h = 6 m above the ground. Block A is moving to the right with an initial vwlocity v = 20 m/s. An elastic collision between the two blocks takes place. What is distance x where block B will hit the ground? S(20) = 1V_B theta = 6 + 0 - 1/2(9.8)t^2 t = 1.01 11.4 m 20 m 33.3 m 36.9 m 2.9 m An ideal spring is hunt? vertically from the ceiling. When a 6.0-kg mass hangs at rest from it, the

Explanation / Answer

final velocity of block A and B are vA and vB are

so for elastic colllision,

20 = vA + vB

vA = 20 - vB   .......(i)

Applying momentum conservation for collision,

5 x 20 + 1 x 0 = 5 x (-vA) + 1vB

vB - 5vA = 100

putting VA from (i)

vB - 5(20 - vB) = 100

6vB = 200

vB = 33.33 m/s

after that block goes h distance vertically down under the acc. due to gravity.

in vertical,

using d = ut + at^2 /2

-h = 0 - gt^2 /2

2 x 6 = 9.8t^2

t = 1.11 sec

in horizontal,

x = vB t = 33.33 x 1.11 = 36.9 m

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