A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to one-fourth its original speed. What is the mass of the second ball?

5/3m

mV=m(-1/4 V)+MV'

1/2 mV^2=1/2m(V^2/16)+1/2MV'^2

form the first equation:
V'=mV(5/4)/M
Put that into the second equation for V', and solve for M

The guy in the top is right !

It’s correct

To find the mass of the second ball, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's denote the mass of the first ball as m1 and the mass of the second ball as m2.

Given:
Before collision:
- Mass of the first ball, m1 = m
- Velocity of the first ball, v1 = initial speed (let's call it v)
- Mass of the second ball, m2 = unknown
- Velocity of the second ball, v2 = 0 (at rest)

After collision:
- Velocity of the first ball (rebounded), v'1 = (1/4)v
- Velocity of the second ball (after collision), v'2 = unknown

Using the conservation of linear momentum, we can write the equation:

(m1 * v1) + (m2 * v2) = (m1 * v'1) + (m2 * v'2)

Now, substitute the known values:

(m * v) + (m2 * 0) = (m * (1/4)v) + (m2 * v'2)

Simplifying this equation will give us the mass of the second ball, m2:

m * v = (m/4) * v + m2 * v'2

mv = (m/4)v + m2v'2

4mv = mv + 4m2v'2

3mv = 4m2v'2

Now, divide both sides of the equation by v'2:

(3mv) / (4v'2) = m2

Finally, cancel out the common factors:

m2 = (3m) / (4v'2)

Therefore, the mass of the second ball, m2, is (3m) / (4v'2).