Question

A ball of mass m makes a head-on elastic collison with a second ball (at rest) and rebounds in the opposite direction with a speed equal to 1/4 it's original speed. What is the mass of the second ball? I would really apprieciate it if ya'll would show steps. I have been working on this for 2 weeks.

Answer 1

For an elastic collision, both momentum and kinetic energy are conserved.

m1 = mass of colliding ball
m2 = mass of ball initially at rest
v1i = initial speed of colliding ball
v1f = final speed of colliding ball
v2i = initial speed of at rest ball = 0
v2f = final speed of at rest ball

v1f = -1/4*v1i = -0.25*v1i

Conservation of momentum:

m1*v1i = -0.25*m1*v1i + m2*v2f

Adding like terms:

1.25*m1*v1i = m2*v2f

Conservation of kinetic energy:

1/2*m1*v1i^2 = 1/2*m*(-v1i/4)^2 + 1/2*m2*v2f^2

Getting rid of the 1/2 factor and adding like terms:

(15/16)*m1*v1i^2 = m2*v2f^2

So now you have 2 useful equations:

1.25*m1*v1i = m2*v2f
(15/16)*m1*v1i^2 = m2*v2f^2

or:

v2f = (1.25*m1*v1i)/m2

(15/16)*m1*v1i^2 = m2*((1.25*m1*v1i)/m2)^2

Use algebra to solve for m2 in terms of v1i and m1

Answer 2

ok...hoow did you get 15/16?

Answer 3

He got 15/16 because you have to take the (-1/4*V1i)^2. You square the (-1/4) to get (1/16) and when you move it over to the other side, you will get m1v1i^2 - (1/16)m1v1i^2 = m2v2f