A 0.68 kg ball traveling at 8.30 m/s collides head-on with a 1.32 kg ball moving in the opposite direction at a velocity of -8.85 m/s. The first ball moves away at -6.28 m/s after the collision. Find the velocity of the second ball.
conserve momentum.
0.68x*8.30 - 1.32*8.85 = -0.68*6.28 +1.32v
To find the velocity of the second ball after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity. Mathematically, momentum (p) is defined as p = m * v, where m is the mass of the object and v is its velocity.
Let's denote the velocity of the second ball after the collision as v2.
The initial momentum of the first ball is given by p1 = m1 * v1, where m1 = 0.68 kg and v1 = 8.30 m/s. The initial momentum of the second ball is given by p2 = m2 * v2, where m2 = 1.32 kg and v2 is the velocity we need to find.
The total momentum before the collision is the sum of the initial momenta of the two balls: p_initial = p1 + p2.
The final momentum of the first ball after the collision is given by p1_final = m1 * v1_final, where v1_final = -6.28 m/s.
According to the principle of conservation of momentum, the total momentum after the collision should be equal to the total momentum before the collision: p_initial = p1_final + p2.
Substituting the values, we get:
m1 * v1 + m2 * v2 = m1 * v1_final + m2 * v2
(0.68 kg * 8.30 m/s) + (1.32 kg * v2) = (0.68 kg * -6.28 m/s) + (1.32 kg * v2)
5.644 kg·m/s + 1.32 kg·v2 = -4.2704 kg·m/s + 1.32 kg·v2
Now, let's simplify the equation by canceling out the terms that appear on both sides:
5.644 kg·m/s = -4.2704 kg·m/s
Next, we'll isolate v2 by moving the terms involving v2 to one side of the equation:
1.32 kg·v2 - 1.32 kg·v2 = -4.2704 kg·m/s - 5.644 kg·m/s
0 = -9.9144 kg·m/s
Since both sides of the equation are equal to zero, this means that the velocity v2 is zero.
Thus, the velocity of the second ball after the collision is 0 m/s.