A 0.060 kg tennis ball, moving with a speed 2.50 m/s, collides head on with a 0.090kg ball initially moving away from it at a speed of 1.15 m/s. Assuming a perfectly elastic collision, what are the speed and direction of each ball after the collision?

Treat bthis as a one-dimensional problem since it is a head-on collision. There is no motion in perpendicular directions.

Problems of this type are most easily done in a coordinate system that moves with the center of mass. Then transform back to laboratory cordinates. You can also treat it only in lab coordinates as two equations in two unknowns, but it takes longer.

The speed of the center of mass of the two balls is Vcm
= (0.06*2.5 + 0.09*1.15)/0.15
= 1.69 m/s

In a coordinate system traveling with the CM, the smaller ball approaches with velocity 2.50-1.69 = 0.81 m/s and the larger ball approaches with a velocity 1.69-1.15 = 0.54 m/s

Both balls reverse direction and keep the same speed in the CM system. The preserves total momentum and kinetic energy.

In laboratory coordinates, the final velocity of the smaller ball is 1.69 - 0.81 = 0.88 m/s (forwards) and the final velocity of the larger ball is
0.54 + 0.81 = 1.35 m/s (also forwards)

thanks :D but where did the 0.15 come from? :) what equation did you use?

0.15 (kg) is the total mass of both balls. You need it in the denominator to get the speed at which the center of mass moves.

My answer had some typos, as usual. "bthis" in the first paragraph and "The" in the next-to-last paragraph should be "this". Note that in the CM coordinate system, total momentum is zero, before and after the collision.

Well, aren't these two balls playing a little game of "I'm Rubber, You're Glue"? Let's see how they bounce off each other!

To solve this head-on collision conundrum, we can use the conservation of momentum. The total momentum before the collision should be equal to the total momentum after the collision. Let's do some math fun!

The initial momentum for the 0.060 kg tennis ball can be calculated as:
Initial momentum = mass × velocity = (0.060 kg) × (2.50 m/s) = 0.15 kg•m/s

The initial momentum for the 0.090 kg ball can be calculated as:
Initial momentum = mass × velocity = (0.090 kg) × (-1.15 m/s) = -0.1035 kg•m/s

Now, the total momentum before the collision is the sum of these two values:
Total initial momentum = (0.15 kg•m/s) + (-0.1035 kg•m/s) = 0.0465 kg•m/s

Since it's a perfectly elastic collision, the total momentum after the collision should also be 0.0465 kg•m/s.

Now we can solve for the final velocities:

Let's denote the velocity of the tennis ball after the collision as V1 and the velocity of the other ball as V2.

Using the conservation of momentum, we can write the equation:
0.060 kg × V1 + 0.090 kg × V2 = 0.0465 kg•m/s

Now to add a dash of algebra to the mix:

0.060 kg × V1 + 0.090 kg × (-1.15 m/s) = 0.0465 kg•m/s

0.060 kg × V1 - 0.1035 kg•m/s = 0.0465 kg•m/s

0.060 kg × V1 = 0.0465 kg•m/s + 0.1035 kg•m/s

0.060 kg × V1 = 0.15 kg•m/s

Finally, we can solve for V1, the final velocity of the tennis ball after the collision:

V1 = 0.15 kg•m/s / 0.060 kg

V1 ≈ 2.5 m/s

So, the speed of the tennis ball after the collision is approximately 2.5 m/s. It's like it's saying, "No matter how hard you hit me, I'll keep bouncing!"

Now let's calculate the final velocity for the other ball, V2:

V2 = (0.0465 kg•m/s - 0.060 kg × V1) / 0.090 kg

V2 ≈ (-0.0015 kg•m/s) / 0.090 kg

V2 ≈ -0.0167 m/s

Ah, poor little ball! It's going backwards with a speed of approximately 0.0167 m/s. It's like saying, "Oops, my bad! I'm just going to sneak away now."

So, after the perfectly elastic collision, the tennis ball has a final velocity of around 2.5 m/s, moving in the same direction it was before, while the other ball has a final velocity of about -0.0167 m/s, moving away in the opposite direction.

And that, my friend, concludes the circus act of these bouncing balls!

To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.

Let's denote the initial speed of the 0.060 kg tennis ball as v1i and the initial speed of the 0.090 kg ball as v2i. After the collision, they will have final speeds v1f and v2f, respectively.

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be written as:

(m1 * v1i) + (m2 * v2i) = (m1 * v1f) + (m2 * v2f),

where m1 and m2 are the masses of the respective balls.

In this case, the collision is perfectly elastic, which means that the kinetic energy is also conserved. So, we can also use the principle of conservation of kinetic energy, which states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Mathematically, this can be written as:

(1/2 * m1 * v1i^2) + (1/2 * m2 * v2i^2) = (1/2 * m1 * v1f^2) + (1/2 * m2 * v2f^2).

Now, let's substitute the given values into these equations:

m1 = 0.060 kg (mass of the tennis ball)
m2 = 0.090 kg (mass of the other ball)
v1i = 2.50 m/s (initial speed of the tennis ball)
v2i = -1.15 m/s (initial speed of the other ball, negative because it's moving away)

Using these values, we can solve the equations to find the final speeds of each ball.

First, let's solve the equation for momentum:

(0.060 kg * 2.50 m/s) + (0.090 kg * -1.15 m/s) = (0.060 kg * v1f) + (0.090 kg * v2f),

0.15 kg·m/s = (0.060 kg * v1f) + (0.090 kg * v2f).

Now, let's solve the equation for kinetic energy:

(1/2 * 0.060 kg * 2.50 m/s^2) + (1/2 * 0.090 kg * (-1.15 m/s)^2) = (1/2 * 0.060 kg * v1f^2) + (1/2 * 0.090 kg * v2f^2).

0.194625 J = (0.030 kg * v1f^2) + (0.04095 kg * v2f^2).

Now we have a system of equations with two unknowns (v1f and v2f). We can solve it using algebra or substitution.

Let's solve the first equation for v2f in terms of v1f:

v2f = (0.15 kg·m/s - 0.060 kg * v1f) / 0.090 kg.

Now substitute this expression for v2f into the second equation:

0.194625 J = (0.030 kg * v1f^2) + (0.04095 kg * ((0.15 kg·m/s - 0.060 kg * v1f) / 0.090 kg)^2).

Now we have a single equation with one unknown (v1f). We can solve it using algebra or numerical methods like solving it with a calculator or software.

Once we find the value of v1f, we can substitute it back into the expression for v2f to find its value.

Finally, we will have the speed and direction of each ball after the collision.