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.
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
ok...hoow did you get 15/16?
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
To solve this problem, we can use the principles of conservation of momentum and kinetic energy.
1. Let's denote the mass of the first ball as m1 and the mass of the second ball as m2.
2. According to the principle of conservation of momentum, the total momentum before and after the collision should be the same. In this case, since the second ball is at rest, we can write the equation as follows:
m1 * v1 = -m1 * v2 (equation 1)
where v1 is the initial velocity of the first ball and v2 is the velocity of the first ball after the collision.
3. The negative sign in the equation accounts for the fact that the two balls are moving in opposite directions after the collision.
4. According to the problem, the speed of the first ball after the collision is equal to 1/4 of its original speed. Since speed is a scalar, the magnitude of the velocity will be 1/4 of the magnitude of the initial velocity. Therefore, we can rewrite equation 1 as:
m1 * v1 = -m1 * (1/4 * v1)
Simplifying, we get:
v1 = -1/4 * v1
5. Now, we can solve for v1:
v1 + 1/4 * v1 = 0
(5/4) * v1 = 0
Therefore, v1 = 0.
This means that the velocity of the first ball before the collision was zero.
Now, knowing that the mass of the first ball is m, and its velocity before the collision is zero, we can use the equation for conservation of kinetic energy.
6. According to the principle of conservation of kinetic energy, the sum of the kinetic energies before and after the collision should be the same. In this case, since the first ball is at rest, the equation is simplified to:
0 + 0 = 1/2 * m * (1/4 * v) ^ 2 (equation 2)
where v is the velocity of the first ball after the collision.
7. Simplifying equation 2, we get:
0 = 1/32 * m * v ^ 2
8. Since the mass of the second ball is denoted as m2, we can rewrite equation 2 as:
0 = 1/32 * m2 * v ^ 2
From equation 2, we see that the mass of the first ball cancels out, meaning that the mass of the second ball does not affect the velocity after the collision. Therefore, we cannot determine the mass of the second ball based on the given information.