A billiard ball rolling across a table at 1.45 m/s makes a head-on elastic colli

Question

A billiard ball rolling across a table at 1.45 m/s makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision when each of the following occurs. (a) The second ball is initially at rest. 1st ball Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s 2nd ball Incorrect: Your answer is incorrect. The correct answer is not zero. m/s (b) The second ball is moving toward the first at a speed of 1.20 m/s. 1st ball m/s 2nd ball m/s (c) The second ball is moving away from the first at a speed of 1.10 m/s. 1st ball m/s 2nd ball m/s

Explanation / Answer

(a)

here by the law of momentum conservation

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

u1 + u2 = v1 + v2

1.45 + 0 = v1 + v2 -----------(i)

for elastic collision:-

v1 - v2 = u2 - u1

v1 - v2 = 0 - 1.45 -----------(ii)

By (i) + (ii) :-

2v1 = 0

v1 = 0 m/s

v2 = 1.45 m/s

(b)

1.45 - 1.20 = v1 + v2

v1 + v2 = 0.25 -----------(i)

and

v1 - v2 = -1.20 - 1.45

v1-v2 = - 2.65 ----------(ii)

from (i) + (ii)

v1 = -1.20 m/s

v2 = 1.45 m/s

(c)

1.45 + 1.1 = v1 + v2

v1 + v2 = 2.55 ----------(i)

v1 - v2 = 1.1 - 1.45

v1 - v2 = -0.35 ---------(ii)

form (i) + (ii) :-

v1 = 1.1 m/s

v2 = 1.45 m/s

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