physics ellastic collision?

Question

A steel ball of mass 0.600 kg is fastened to a cord that is 60.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal . At the bottom of its path, the ball strikes a 3.00 kg steel block initially at rest on a frictionless surface. The collision is elastic.


(a) Find the speed of the ball just after collision.
m/s
(b) Find the speed of the block just after collision.
m/s

I already tried this a few times and the speed of the ball after collision is neither zero, or 3.43,(which is what I get from conservation of energy calculations.

Answer 1

♠ the total energy at the collision is pot energy
E =m1*gh, where m1=0.6 kg, h=0.6 m;
♣ thus kin energy of the ball just before collision
E=0.5m1*v^2; thence v=√(2gh);
♦ now after collision: momentum conserv law says
m1*v1+m2*v2 =m1*v, where m2=3kg, v1 is speed of the ball, v2 is speed of the block; thence v2=m1*(v-v1)/m2;
and energy conserv law says
0.5m1*v1^2 +0.5m2*v2^2 = E; or;
m1*v1^2 +m2*{m1*(v-v1)/m2}^2 = 2m1*gh; or;
v1^2 +(m1/m2)*{v^2 -2v*v1 +v1^2} =2gh; or;
♥ (m1+m2)*v1^2 –2m1*√(2gh)*v1 –(m2-m1)*2gh =0;
hence v1=(m1*√(2gh) +S)/(m1+m2), where
S =±√(m1^2 *2gh +(m2^2-m1^2) *2gh) = ±m2*√(2gh);
And v1=(m1±m2)* √(2gh) /(m1+m2);
the solution with plus had been known before collision;
▼thus after collision
v1= (m1-m2)*√(2gh) /(m1+m2) = -2.2862 m/s; minus means rebounding!
And v2= v2=m1*(v-v1)/m2 (see ♦); or;
v2= 2m1*√(2gh) /(m1+m2) = +1.1431 m/s; plus means forward;

Answer 2

Hey, I've got the idea, the speed of the ball just after the collision could be zero, but, not necesary, because we need to remember that the linear momentum is constant before and after the collision :

0.6v = 0.6v' + 3v''

Now, if we consider v', the speed of the ball just after the collision zero, then :

0.6v = 3v''

but, do you remember :

e = v'' + v' / v

where e =1

v'' - v' = v

and, what about 60 cm ?

if the ball falling from the rest, then Vo = 0

and Vf^2 = 0 +2*9.8*60 / 100 = 11.76 m/s >>> v of the ball before the collision, then :

11.76*0.6 = 0.6v' + 3*v''

v'' - v' = 11.76

two equations with two variables, then :

v' = 3.92 >>> speed of the ball after the collision

v'' = 15.68 >>> speed of the block after the collision

Look, this is like a special case, considering that both of the balls are moving after the colission, and that the first ball is falling from 60 cm.

g = 9.8 m/s^2

Answer 3

(a) zero

(b) m1v1 (of ball just before collision) = m2v2 (of block just after collision, and for the life of the planet if there is no friction and the surface completely circles the earth).

So v2 = m1v1 / m2 = 0.600/3.00 * v1 = 0.200 * v1.

You solve for v1.