Ball A with a mass of 3 kg moving 15 m/s east collides ball B (a 5 kg ball movin

Question

Ball A with a mass of 3 kg moving 15 m/s east collides ball B (a 5 kg ball moving 6 m/s west). If the coefficient of restitution was equal to 1, then after impact, the velocity of ball A would be ________________ and the momentum of ball B would be ______________:

A. 10 m/s west; 45 kg*m/s east

B. 15 m/s west; 30 kg*m/s east

C. 15 m/s east; 30 kg*m/s west

D. 5 m/s east; 3 m/s west

E. 5 m/s west; 3 m/s east

A. 10 m/s west; 45 kg*m/s east

B. 15 m/s west; 30 kg*m/s east

C. 15 m/s east; 30 kg*m/s west

D. 5 m/s east; 3 m/s west

E. 5 m/s west; 3 m/s east

Explanation / Answer

The velocity of the ball A after collision is

VA2 = ((mA-mB)/ (mA+mB))VA1 + ( 2mB / mA+mB) VB1

VA2 = ((3-5/3+5) 15 )+ ((2 X5)/(3+5))(-6)

=-3.75-7.5 = -11.25M/S (~10m/s west)

-ve sign represents opposite to the original direction

the velocity of the ball B after collision is

VB2 = ((mB-mA)/ (mA+mB))VB1 + ( 2mA / mA+mB) VA1

VB2 = ((5-3)/(5+3)) (-6)+ ((2X3)/(3+5))(15)

            = -1.5+11.25 =9.75 (WEST)

Momentum of the Ball B is P= 5X9.75 =48.75kgm/s (~45kg m/s) east

Correct answer is A

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