physics help?

Question

The pendulum bobs in the figure below are made of soft clay so that they stick together after impact. The mass of bob A is half that of bob B. Bob B is initially at rest. What is the ratio of the kinetic energy of the combined bobs, just after impact, to the kinetic energy of bob A just before impact?

Answer 1

Momentum is *always* conserved. Bob A has mass a and bob B has mass 2a. If bob A has velocity v then, just after the impact, bobs A nad B together have velocity v/3. Since kinetiic energy is 1/2mv², the kinetic energy of the two bobs together will be 1/3 of the kinetic energy of bob A alone before the impact.


Doug

Answer 2

It is 1:3, KE of BobA+Bob B to that of Bob A

let mass of Bob A = 2x
Mass of Bob B = 4x

K:E of Bob A before impact
1/2(2x)v^2
=xv^2

Total mass after impact= 2x+4x=6x

momentum before and after impact are equal.

6x*u= 2x*v

u=1/3v (u=velocity of both bobs after impact)

K:E of Both bobs after impact

1/2(6x)(1/3v)^2
=3x(1/9v^2)
=1/3xv^2

Ratio of K:E of Bob A + Bob B to that of Bob A

1/3xv^2 / xv^2

=1/3 / 1
=1/3 *3 / 1*3
=1/3

1 ratio 3

1:3