7.78 roblem 7.78 Two balls, of masses mA =45 gand mB-61 g.are pulled away to a 6

Question

7.78 roblem 7.78 Two balls, of masses mA =45 gand mB-61 g.are pulled away to a 66 angle with the vertical and Determine the velocity of ball A before impact as shown in (Figure 1). The lighter ball is Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the velocity is to the right negative value if the direction of the velocity is to the left of 1 Part B Determine the velocity of ball A after the elastic colision. Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the velocity is to the right negative value if the direction of the velocity is to the left. 35 cm Submit My Answers Give Up

Explanation / Answer

a)

The height of A at 66° is 35 cm - x, and x = 35*cos66° = 14.236 cm

A = 0.14236 m

Now set potential energy equal to kinetic energy:

mgh = 1/2 mv^v^2

v^2 = 2gh

v = sqrt(2gh)

v = sqrt(2*9.81*0.14236)

v = 1.671 m/s velocity of A before it hits B

The velocities of both balls after collision are

v1' = (m1v1 + m2(2v2-v1))/(m1+m2)

v1' = [(0.045*1.671) + 0.061(- 1.671)] / 0.106

v1' = - 0.2522 m/s = velocity of A ..........Ans(b)

v2' = (m2*v2 + m1(2v1 - v2)) / (m1+m2)

v2' = 0.045(2*1.671) / 0.106

v2' = 1.4188 m/s = velocity of B ..........Ans(c)

for the height they reach use again

mgh = 1/2 mv^2

h = v^2/(2g)

height of A = 0.2522^2 / (2*9.81)

height of A = 0.3242 cm ............Ans(d)

height of B = 1.4188^2/(2*9,81)

height of B = 10.26 cm ....Ans(e)

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