A ball with mass m = 0.210 kg and kinetic energy K1 = 2.97 J collides elastically with a second ball of the same mass that is initially at rest. After the collision, the first ball moves away at an angle of = 30.6° with respect to the hori- zontal, as shown in the figure. What is the kinetic energy of the first ball after the collision?

Please step by step

I only got the the first V of M1 that it is 5.31 m/s :/

Well, it looks like you've already gotten started on the problem! Don't worry, I'm here to help you with the rest of it step by step.

Let's start by finding the initial velocity of the first ball (let's call it V1i). We can use the formula for kinetic energy:

K = (1/2) * m * V^2

where K is the kinetic energy, m is the mass, and V is the velocity.

Given that K1 = 2.97 J, we can plug in the values:

2.97 J = (1/2) * 0.210 kg * V1i^2

Now let's solve for V1i:

2.97 J = 0.105 kg * V1i^2

Divide both sides by 0.105 kg:

V1i^2 = 2.97 J / 0.105 kg

V1i^2 = 28.2857 m^2/s^2

Now take the square root of both sides to find V1i:

V1i = √(28.2857 m^2/s^2)

V1i ≈ 5.31 m/s

Awesome! You already found the initial velocity of the first ball correctly. Now, let's move on to finding the final velocity of the first ball (V1f) after the collision.

Since the collision is elastic, kinetic energy is conserved. So, we can use the formula:

K1i + K2i = K1f + K2f

where K1i and K2i are the initial kinetic energies of the first and second balls respectively, and K1f and K2f are their final kinetic energies.

However, we are given that the second ball is initially at rest, so K2i = 0.

Therefore, the equation simplifies to:

K1i = K1f + K2f

Plugging in the values:

2.97 J = K1f + K2f

Now, let's find the final velocity of the first ball. We know that kinetic energy can also be expressed as:

K = (1/2) * m * V^2

Given that both balls have the same mass (0.210 kg), we can write:

K1f = (1/2) * 0.210 kg * V1f^2

Plugging this into our equation:

2.97 J = (1/2) * 0.210 kg * V1f^2 + K2f

Since K2f depends on the final velocity of the second ball (V2f), we can't solve for K2f directly. However, we can use the principle of conservation of momentum to find a relationship between the two velocities.

The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. In this case, since the second ball is initially at rest, the momentum before the collision is simply the momentum of the first ball:

p1i = m1 * V1i

where p1i is the initial momentum, m1 is the mass of the first ball, and V1i is its initial velocity.

After the collision, the first ball moves away at an angle of 30.6° with respect to the horizontal. The horizontal component of its velocity (V1fx) can be found using trigonometry:

V1fx = V1f * cos(θ)

where θ is the angle (30.6°).

Now let's substitute the values in and solve for V1f:

V1f * cos(30.6°) = V1i

V1f = V1i / cos(30.6°)

Plugging in the values:

V1f ≈ 5.31 m/s / cos(30.6°)

V1f ≈ 6.13 m/s

Fantastic! Now we know the final velocity of the first ball. Let's go back to our equation for kinetic energy and find K2f.

2.97 J = (1/2) * 0.210 kg * (6.13 m/s)^2 + K2f

2.97 J = 0.105 kg * (37.5169 m^2/s^2) + K2f

2.97 J = 3.939 m^2/s^2 + K2f

Subtracting 3.939 m^2/s^2 from both sides:

2.97 J - 3.939 m^2/s^2 = K2f

-0.969 J = K2f

Now we know that the final kinetic energy of the second ball (K2f) is equivalent to -0.969 J.

But wait! Since kinetic energy cannot be negative, we conclude that the second ball must have moved in the opposite direction as we assumed. In other words, the second ball moves towards the left after the collision.

To determine the kinetic energy of the first ball after the collision, we can use the conservation of momentum and the conservation of kinetic energy.

1. Conservation of momentum:
Before the collision, the total momentum is zero since the second ball is initially at rest. Therefore, the total momentum after the collision must also be zero. We can write this as:

m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final

Since the second ball is initially at rest, its initial velocity (v2_initial) is 0. The equation simplifies to:

m1 * v1_initial = m1 * v1_final

2. Conservation of kinetic energy:
The kinetic energy before and after the collision must also be conserved. Thus, we can write:

(1/2) * m1 * (v1_initial)^2 + (1/2) * m2 * (v2_initial)^2 = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final)^2

Since the second ball is initially at rest, its initial kinetic energy (K2_initial) is 0. The equation simplifies to:

(1/2) * m1 * (v1_initial)^2 = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final)^2

Now, we can substitute the value of v1_initial (which you found to be 5.31 m/s) into the equations and solve for v1_final.

m1 * v1_initial = m1 * v1_final

0.210 kg * (5.31 m/s) = 0.210 kg * v1_final

v1_final = 5.31 m/s

Now, substitute the value of v1_final into the second equation:

(1/2) * m1 * (5.31 m/s)^2 = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final)^2

(1/2) * 0.210 kg * (5.31 m/s)^2 = (1/2) * 0.210 kg * (5.31 m/s)^2 + (1/2) * 0.210 kg * (v2_final)^2

Simplify and solve for (v2_final)^2:

(1/2) * 0.210 kg * (v2_final)^2 = 0

(v2_final)^2 = 0

Therefore, the second ball does not move after the collision, and its final velocity is 0 m/s.

Finally, substitute the values of v1_final = 5.31 m/s and v2_final = 0 m/s into the equation for the kinetic energy of the first ball after the collision:

K1_final = (1/2) * m1 * (v1_final)^2
K1_final = (1/2) * 0.210 kg * (5.31 m/s)^2

Now, calculate the final kinetic energy:

K1_final = 0.5 * 0.210 kg * (5.31 m/s)^2
K1_final = 0.55814 J

Therefore, the kinetic energy of the first ball after the collision is approximately 0.55814 J.

To find the kinetic energy of the first ball after the collision, we need to break down the problem into smaller steps:

1. Calculate the initial velocity of the first ball before the collision:
- Given: mass of the ball (m) = 0.210 kg, kinetic energy before collision (K1) = 2.97 J
- We know that kinetic energy (K) can be calculated using the formula K = 0.5 * m * v^2, where v is the velocity.
- Rearranging the formula, we have v = sqrt(2K / m).
- Substituting the given values, we get v = sqrt(2 * 2.97 J / 0.210 kg) = 5.31 m/s.

2. Determine the velocity components of the first ball after the collision:
- The first ball moves away at an angle of 30.6° with respect to the horizontal.
- We need to find the horizontal and vertical components of its velocity.
- The horizontal component (Vx) can be calculated using the formula Vx = v * cos(θ).
- Here, v is the magnitude of the velocity (5.31 m/s) and θ is the angle (30.6°).
- Substituting the values, we get Vx = 5.31 m/s * cos(30.6°) = 4.59 m/s.
- The vertical component (Vy) can be calculated using the formula Vy = v * sin(θ).
- Substituting the values, we get Vy = 5.31 m/s * sin(30.6°) = 2.66 m/s.

3. Calculate the kinetic energy of the first ball after the collision:
- Given that the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
- The initial kinetic energy of the first ball is given as K1 = 2.97 J, and we need to find the final kinetic energy.
- The final kinetic energy can be found using the formula K = 0.5 * m * v^2, where v is the final velocity.
- We need to calculate the magnitude of the final velocity (v_final) of the first ball after the collision using the horizontal and vertical components.
- The magnitude of the velocity (v) can be calculated using the formula v = sqrt(Vx^2 + Vy^2).
- Substituting the values, we get v = sqrt((4.59 m/s)^2 + (2.66 m/s)^2) = 5.37 m/s.
- Now, we can find the final kinetic energy using the formula K = 0.5 * m * v_final^2.
- Substituting the values, we get K = 0.5 * 0.210 kg * (5.37 m/s)^2 = 2.14 J.

Therefore, the kinetic energy of the first ball after the collision is 2.14 J.