A softball of mass 0.200 kg that is moving with a speed of 7.8 m/s collides head

Question

A softball of mass 0.200 kg that is moving with a speed of 7.8 m/s collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 3.1 m/s (please nswer all parts!! see photo)

Problem 7.26 Part A A softball of mass 0.200 kg that is moving with a speed of 7.8 m/s collides head-on and elastically with another ball initially at rest. Afterward the incoming softbal bounces backward with a speed of 3.1 m/s Calculate the velocity of the target ball after the collision Express your answer using two significant figures m/s Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Part B Calculate the mass of the target ball Express your answer using two significant figures m= Submit My Answers Give Up Provide Feedback Conti

Explanation / Answer

m1 = mass of softball = 0.2 kg

m2 = mass of target ball = m

V1i = initial velocity of softball before collision = 7.8 m/s

V1f = final velocity of softball after collision = - 3.1 m/s

V2i = initial velocity of target ball before collision = 0 m/s

V2f = final velocity of target ball after collision = v

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

(0.2) (7.8) + m (0) = (0.2) (-3.1) + m v

mv = 2.18                     

m = 2.18/v                               eq-1

using conservation of kinetic energy

(0.5) m1 v21i + (0.5) m2 v22i = (0.5) m1 v21f + (0.5) m2 v22f

m1 v21i + m2 v22i = m1 v21f + m2 v22f

(0.2) (7.8)2 + (m) (0)2 = (0.2) (-3.1)2 + m v2

(0.2) (7.8)2 = (0.2) (-3.1)2 + (2.18/v) v2                       using eq-1

v = 4.7 m/s

using eq-1

m = 2.18/v = 2.18 /4.7 = 0.46 kg

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