Momentum of two balls physics

[chazgurl4life]

Two balls, of masses mA = 32 g and mB = 80 g are suspended as shown in Figure 7-44. The lighter ball is pulled away to a 60° angle with the vertical and released.a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)


a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)

i have been using the formula V=M+m
---- X (2gl(1-cos of the angle)^1/2
m


so when i put it all together i get :

V= .08kg+.032Kg
---------- X (2*9.8*.3m*1-cos60 degrees)^1/2
.032 kg

V=3.5 X(2.94)^1/2
V=6 m/s

but apparently this answer is wrong ..can anyone help ??

 

[Chi Meson]

You solved too far. The answer is in your question.(modified slightly with brackets).

a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)

[2gl(1-cos of the angle)]^1/2

 

[kyle8921]

I would like to clarify that the formula you have used is incorrect. The correct formula for calculating the velocity of the lighter ball before impact is V = (mB / mA) X (gL(1-cosθ))^1/2, where mB is the mass of the heavier ball, mA is the mass of the lighter ball, g is the acceleration due to gravity, L is the length of the string, and θ is the angle the lighter ball is pulled to.

Using this formula, we can calculate the velocity of the lighter ball before impact as follows:

V = (0.08 kg / 0.032 kg) X (√(2 X 9.8 X 0.3 X (1-cos60 degrees)))
V = 2.5 X (√(5.88))
V = 2.5 X 2.42
V = 6.05 m/s

Therefore, the velocity of the lighter ball before impact is approximately 6.05 m/s, which is slightly different from your calculated value of 6 m/s. This could be due to rounding errors or a slight difference in the values used in the formula. It is important to use the correct formula and accurate values in order to obtain the correct answer in physics.