Verified answer

@ peter: yo... what have you been smoking man ???

m₁ = mass of the ball = 0.600 kg

m₂ = mass of the block = 4.60 kg

L = length of the cord = 0.700 m

g = acceleration by gravity = 9.81 m/s²

The potential energy of the ball at its initial position is:

PE = m₁ g L

That is also the kinetic energy of the ball at its lowest point,

so its speed at that point is:

KE = PE

m₁ v² / 2 = m₁ g L

v = √( 2 g L )

v = 3.7059 m/s

The equations for elastic collision are:

(see http://en.wikipedia.org/wiki/Elastic_collision )

m₁×u₁ + m₂×u₂ = m₁×v₁ + m₂×v₂

and

v₁ - v₂ = u₂ - u₁

where

m₁ = mass of the ball = 0.600 kg

u₁ = speed of the ball just before the collision = v = 3.706 m/s

v₁ = speed of the ball just after the collision = ?

m₂ = mass of the block = 4.60 kg

u₂ = speed of the bloc just before the collision = 0 m/s

v₂ = speed of the bloc just after the collision = ?

So:

(0.6 kg)×(3.706 m/s) + (4.6 kg)×(0 m/s) = (0.6 kg)×v₁ + (4.6 kg)×v₂

(2.224 kgm/s) = (0.6 kg)×v₁ + (4.6 kg)×v₂ . . . . . . . . equation 1

v₁ - v₂ = (0 m/s) - (3.706 m/s)

v₂ = v₁ + (3.706 m/s) . . . . . . . . . . . . . . . . . . . . . . . equation 2

Substitute equation 2 into equation 1.

(2.224 kgm/s) = (0.6 kg)×v₁ + (4.6 kg)×(v₁ + (3.706 m/s))

(2.224 kgm/s) = (0.6 kg)×v₁ + (4.6 kg)×v₁ + (17.05 kgm/s)

(-14.82 kgm/s) = (5.2 kg)×v₁

v₁ = -2.85 m/s

Substitue that value into equation 2.

v₂ = (-2.85 m/s) + (3.706 m/s)

v₂ = 0.855 m/s

You now found the VELOCITY of the ball and block just after the collision.

The opposite direction of their motion is indicated by their opposite sign.

ANSWER:

the SPEED of the objects after the collision is

ball : 2.85 m/s

block : 0.855 m/s

Sorry however i cant provide an explanation for with values coz it is going to take too much time to calculate . I will most effective tell u the process . Us mgh = exchange in kinetic power of ball for two positions of the , horizontal and vertical . So u can get speed of ball simply earlier than placing . Now use conservation momentum . So now u will hav one eqn. With 2 unknwns . Now as collision is elastic complete vigour earlier than and after the collision dame . So ok.E. Before collision = okay.E. After collision [i.E. Ke of both the bodies]. So now u will jav amother eqn. With equal 2 unkwns as above . Resolve these 2 eqns. Amf u will get required ans.

sorry but i cant explain with values coz it will take too much time to calculate . i will only tell u the procedure . us mgh = change in kinetic energy of ball for 2 positions of the , horizontal and vertical . so u can get velocity of ball just before striking . now use conservation momentum . so now u will hav one eqn. with 2 unknwns . now as collision is elastic total energy before and after the collision dame . so k.e. before collision = k.e. after collision [i.e. ke of both the bodies]. so now u will jav amother eqn. with same 2 unkwns as above . solve these 2 eqns. amf u will get required ans.

speed of the ball (when released) is given by

1/2mv^2=mgh

v0=sqrt(2gh)

v0=3,7 m/s

the speed at the bottom of its path is

v=sqrt(v0^2+2gh)

v=5.2 m/s

the speed of the ball and that of the block after collision are given by

m1v1i=m1v1f+m2v2f

1/2mv1i^2=1/2m1v1f^2+1/2m2v2f*2

then

v1f=[(m1-m2)v1i]/(m1+m2)=-4 m/s [ball]

v2f=[2m1v1i]/(m1+m2)=1.2 m/s [block]