Two balls, of masses m A = 45 g and m B = 70 g , are suspended as shown in (Figu

Question

Two balls, of masses mA = 45 g and mB = 70 g , are suspended as shown in (Figure 1) . The lighter ball is pulled away to a 66? angle with the vertical and released.

Part A
Determine the velocity of ball A before impact.

Part B

Determine the velocity of ball A after the elastic collision.

Part C

Determine the velocity of ball B after the elastic collision.

Part D

Determine the maximum height of ball A after the elastic collision.

Part E

Determine the maximum height of ball B after the elastic collision.

Explanation / Answer

at A

potential energy PEA = mA*g*L*(1-cos66)

at the lowest point before impact KEA = 0.5*mA*vA^2

from energy conservation

KEA = PEA

0.5*mA*vA^2 = mA*g*L*(1-cos66)

0.5*vA^2 = g*L*(1-cos66)

0.5*vA^2 = 9.8*0.35*(1-cos66)

vA1 = 2.02 m/s

++

part(B)

frommomentum conservation

mA*vA1 + mB*vB1 = mA*vA2 + mB*vB2

(0.045*2.02) + (0.07*0) = (0.045*vA2) + (0.07*vB2)

from energy conservation

0.5*mA*vA1^2 + 0.5*mB*vB1^2 = 0.5*mA*vA2^2 + 0.5*mB*vB2^2

0.5*0.045*2.02^2 + 0.5*0.07*0 = 0.5*0.045*vA2^2 + 0.5*0.07*vB2^2

solving the two equations

part(B)

velocity of A after collision = vA2 = -0.44 m/s

part(c)

velocity of B after collision = vB2 = 1.58 m/s

part(D)

hA = vA2^2/2g = 0.44^2/(2*9.8) = 0.0098 m

part(E)

hB = vB^2/2g = 1.58^2/(2*9.8) = 0.13 m

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