Help me pls?

Question

Two balls, of masses mA = 24 g and mB = 76 g are suspended as shown in Figure 7-44. The lighter ball is pulled away to a 60° angle with the vertical and released. Length of string is 30 cm.

What is the velocity of Ball B after collision.

The velocity of Ball A before collision is 1.71.
Va=square root of 2(Distance X g X D cos(angle) X gravity)
=1.71
Velocity of Ball A after collision -.8892.
Ma-Mb divided by Ma+Mb x 1.71
=-.8892

Again we are looking for:

What is the velocity of Ball B after collision.

Answer 1

I find that vB is simply vA1 + vA2 = 1.7426... - 0.8916... m/s = 0.8230...m/s.

Since you don't give a diagram, etc., it's a little hard to visualize. However, my calculation of the speed of A on impact is:

vA1^2 = 2 (15/100) (9.8)m/s^2 = 2.94m/s^2

==> vA1 = 1.7146... m/s (agreeing with your result, but with more digits).

Since you tell us nothing else, I shall assume that the impact is elastic, i.e. that ball A bounces off ball B with exactly the speed with which it approached it. I agree that conservation of linear momentum at the moment of impact then implies that:

vA2 = [(mA - mB) / (mA + mB)] vA1 = - 0.8916... m/s.

This is very close to your result (and indeed I found that I got your result if I rounded my value 1.7146... to your 1.71). So basically, I agree with you completely so far, with only a slight numerical difference.

But vB is simply vA1 + vA2 = 1.7426... - 0.8916... m/s = 0.8230...m/s.

Live long and prosper.