A 2.20-kg ball, moving to the right at a velocity of +1.27 m/s on a frictionless table, collides head-on with a stationary 7.20-kg ball. Find the final velocities of (a) the 2.20-kg ball and of (b) the 7.20-kg ball if the collision is elastic. (c) Find the magnitude and direction of the final velocity of the two balls if the collision is completely inelastic.

Question

I solved part b, I need help with part a if possible, cause if I solve a I could easily solve part c

Answer 1

sorry I solved part C, I need help with a and b pls.

Answer 2

ok I got the answer:
a) Vf1=(m1-m2/m1+m2)*(Vi1)
B) Vf2=[(2(m1)/m1+m2](Vi1)

Answer 3

Final Velocity of 2.2kg ball:

Vf = (V_i *(m1-m2) + (2m2*v_i2)/(m1+m2)

v_i =1.27
m1=2.20
m2=7.2
v_i2 = 0 (because it was a rest initially)

Vf=(1.27*(2.2-7.2)+(2*7.2*0)/(1.27+7.2))
=-6.35

For final Velocity of 2nd object:
Vf = (V_i2 *(m2-m1) + (2m1*v_i/(m1+m2)
Vf=(0*(7.2-2.2)+(2*2.2*1.27)/(1.27+7.2))
Vf=.66

So answer is
-6.35 and .66

Someone check my method and math.

Answer 4

yeah you are right, ignore my post.