A ball with a mass of 4 kg and a velocity of 10 m/s strikes an 8kg ball that is initially at rest. If the collision is perfectly elastic, what are the velocities of the 4 kg and 8 kg ball after the collision?
To find the velocities of the balls after the collision, we will use the principles of conservation of momentum and kinetic energy.
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, it can be written as:
(đ1 * đŁ1) + (đ2 * đŁ2) = (đ1 * đŁ1â˛) + (đ2 * đŁ2â˛)
Where:
m1 = mass of the first ball
v1 = initial velocity of the first ball
m2 = mass of the second ball
v2 = initial velocity of the second ball
v1' = final velocity of the first ball
v2' = final velocity of the second ball
In this case:
m1 = 4 kg
v1 = 10 m/s
m2 = 8 kg
v2 = 0 m/s (as it is initially at rest)
Since the collision is perfectly elastic, kinetic energy is conserved as well. The kinetic energy before and after the collision can be represented using the equation:
(1/2 * m1 * v1^2) + (1/2 * m2 * v2^2) = (1/2 * m1 * v1'^2) + (1/2 * m2 * v2'^2)
Plugging in the given values, we get:
(1/2 * 4 * 10^2) + (1/2 * 8 * 0^2) = (1/2 * 4 * v1'^2) + (1/2 * 8 * v2'^2)
Now, let's solve for v1' and v2'.
(1/2 * 4 * 10^2) = (1/2 * 4 * v1'^2) + (1/2 * 8 * v2'^2)
(2 * 100) = (2 * v1'^2) + (4 * v2'^2)
200 = 2v1'^2 + 4v2'^2
Divide the equation by 2:
100 = v1'^2 + 2v2'^2
We have two equations:
(4 * 10) + (0) = (4 * v1') + (8 * v2') (From the conservation of momentum)
100 = v1'^2 + 2v2'^2 (From the conservation of kinetic energy)
Now we have a system of equations. By solving them simultaneously, we can find the values of v1' and v2'.
4 * 10 = 4 * v1' + 8 * v2' => 40 = 4v1' + 8v2'
100 = v1'^2 + 2v2'^2
To simplify the equations, let's divide both sides of the first equation by 4:
10 = v1' + 2v2' (Equation 1)
Now substitute Equation 1 in Equation 2:
100 = (v1' + 2v2')^2 + 2v2'^2
Expanding the equation:
100 = v1'^2 + 4v2'^2 + 4v1'v2' + 2v2'^2
Combine like terms:
100 = v1'^2 + 6v2'^2 + 4v1'v2'
Now substitute the value of v1' from Equation 1:
100 = (10-2v2')^2 + 6v2'^2 + 4(10-2v2')(v2')
Expand and simplify:
100 = 100 - 40v2' + 4v2'^2 + 6v2'^2 + 40v2' - 8v2'^2
Combine like terms:
100 = 100 - 2v2'^2
-2v2'^2 = 0
v2'^2 = 0
Therefore, the final velocity of the 8 kg ball after the collision is 0 m/s.
Now, substitute this value in Equation 1:
10 = v1' + 2 * 0
v1' = 10 m/s
Therefore, the final velocity of the 4 kg ball after the collision is 10 m/s.