A 120 g ball moving to the right at 4.2 m/s catches up and collides with a 450 g ball that is moving to the right at 1.2 m/s. If the collision is perfectly elastic, what are the speed and direction of each ball after the collision?
.12(v1)+.42(v1+3)=
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
Step 1: Calculate the initial momentum of each ball.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v). Therefore, the initial momentum of the 120 g ball is:
p1 = m1 * v1
= 0.120 kg * 4.2 m/s
= 0.504 kg·m/s
Similarly, the initial momentum of the 450 g ball is:
p2 = m2 * v2
= 0.450 kg * 1.2 m/s
= 0.54 kg·m/s
Step 2: Apply conservation of momentum.
In a perfectly elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Therefore:
p1 + p2 = p1' + p2'
Where p1' and p2' are the final momenta of the balls.
Substituting the initial momenta, we have:
0.504 kg·m/s + 0.54 kg·m/s = p1' + p2'
Step 3: Apply conservation of kinetic energy.
In a perfectly elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Therefore:
(1/2) m1 v1^2 + (1/2) m2 v2^2 = (1/2) m1 (v1')^2 + (1/2) m2 (v2')^2
Where v1' and v2' are the final velocities of the balls.
Substituting the initial masses and velocities, we have:
(1/2) * 0.120 kg * (4.2 m/s)^2 + (1/2) * 0.450 kg * (1.2 m/s)^2 = (1/2) * 0.120 kg * (v1')^2 + (1/2) * 0.450 kg * (v2')^2
Step 4: Solve the equations.
So we have two equations:
0.504 kg·m/s + 0.54 kg·m/s = p1' + p2'
(1/2) * 0.120 kg * (4.2 m/s)^2 + (1/2) * 0.450 kg * (1.2 m/s)^2 = (1/2) * 0.120 kg * (v1')^2 + (1/2) * 0.450 kg * (v2')^2
Simplifying the equations, we get:
1.044 = v1' + v2'
0.0792 + 0.324 = 0.06 * (v1')^2 + 0.135 * (v2')^2
We now have a system of two equations. Solving this system will give us the final velocities of the balls after the collision.