A 0.280 kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.

(a) What is the mass of the second ball?

(b) What fraction of the original kinetic energy (ΔKE/KE) gets transferred to the second ball?
I'm not sure where to start this problem. Any help is appreciated!

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

(a) To find the mass of the second ball, we need to use the principle of conservation of momentum. According to this principle, the total momentum before the collision equals the total momentum after the collision.

So, we can write the equation:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')

where:
m1 and m2 are the masses of the croquet ball and the second ball, respectively.
v1 and v2 are the initial velocities of the croquet ball and the second ball, respectively.
v1' and v2' are the final velocities of the croquet ball and the second ball, respectively.

In this scenario, the second ball is initially at rest, so v2 = 0. Also, the second ball moves off with half the original speed of the first ball, so v2' = 0.5 * v1.

The equation becomes:
(m1 * v1) + (m2 * 0) = (m1 * v1') + (m2 * 0.5 * v1)

Since the second term on the left side is zero, the equation simplifies to:
m1 * v1 = m1 * v1' + 0.5 * m2 * v1.

Dividing both sides of the equation by v1:
m1 = m1 * (v1' + 0.5 * m2)

Finally, dividing both sides by m1:
1 = v1' + 0.5 * m2

Rearranging the equation, we get:
m2 = (1 - v1') / 0.5

Now, we know that m1 = 0.280 kg and can substitute the values to calculate m2.

(b) To find the fraction of the original kinetic energy transferred to the second ball, we can use the principle of conservation of kinetic energy. According to this principle, the total kinetic energy before the collision equals the total kinetic energy after the collision.

The equation for kinetic energy is given by:
KE = (1/2) * m * v^2

The initial kinetic energy of the croquet ball is given by:
KE1 = (1/2) * m1 * v1^2

The final kinetic energy of the second ball is given by:
KE2 = (1/2) * m2 * v2'^2

The fraction of the original kinetic energy transferred to the second ball can be calculated as:
ΔKE/KE = (KE2 - KE1) / KE1

Now, substituting the values of m1, m2, v1, and v2' into the equations, we can calculate the fraction of the original kinetic energy transferred to the second ball.

I hope this explanation helps you understand how to approach this problem. Let me know if you need any further assistance!

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

Let's consider the following information:

Let's assume the initial velocity of the first ball is v1, and its mass is m1 (given as 0.280 kg).
The velocity of the second ball after the collision is v2, and its mass is m2 (unknown).
The velocity of the second ball after the collision is half the original velocity of the first ball.

Now let's solve the problem step by step:

(a) What is the mass of the second ball?

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

The momentum of a ball is given by the product of its mass and velocity, so the equation is:

(mass1 x velocity1) + (mass2 x velocity2) = (mass1 x final velocity1) + (mass2 x final velocity2)

Considering that the second ball is initially at rest (velocity2 = 0), and the final velocity of the first ball is v1/2, we can rewrite the equation as:

(mass1 x velocity1) = (mass1 x v1/2) + (mass2 x 0)

Substituting the known values:

(0.280 kg x v1) = (0.280 kg x (v1/2)) + (mass2 x 0)

Cancellation and further simplification:

0.280 kg x v1 = 0.140 kg x v1

Since v1 is on both sides of the equation, it cancels out, leaving us with:

0.280 kg = 0.140 kg + mass2

Simplifying further:

mass2 = 0.280 kg - 0.140 kg
mass2 = 0.140 kg

Therefore, the mass of the second ball is 0.140 kg.

(b) What fraction of the original kinetic energy (ΔKE/KE) gets transferred to the second ball?

The kinetic energy of an object is given by the equation:

KE = (1/2) x mass x velocity^2

Let's calculate the initial kinetic energy (KE1) of the first ball and the final kinetic energy (KE2) of the second ball.

KE1 = (1/2) x mass1 x (velocity1)^2
KE1 = (1/2) x 0.280 kg x (v1)^2

KE2 = (1/2) x mass2 x (velocity2)^2
KE2 = (1/2) x 0.140 kg x (v1/2)^2

Now let's calculate the fraction of the original kinetic energy (ΔKE/KE) transferred to the second ball.

ΔKE = KE2 - KE1

ΔKE/KE = (ΔKE/KE1) = (KE2 - KE1)/KE1

Substituting the values:

(ΔKE/KE) = ((1/2) x 0.140 kg x (v1/2)^2 - (1/2) x 0.280 kg x (v1)^2) / ((1/2) x 0.280 kg x (v1)^2)

Simplifying further:

(ΔKE/KE) = ((1/4) x 0.140 kg x v1^2 - (1/2) x 0.280 kg x v1^2) / ((1/2) x 0.280 kg x v1^2)

(ΔKE/KE) = (0.035 kg x v1^2 - 0.140 kg x v1^2) / (0.140 kg x v1^2)

(ΔKE/KE) = (-0.105 kg x v1^2) / (0.140 kg x v1^2)

(ΔKE/KE) = -0.75

The negative sign indicates that the second ball received less kinetic energy than the first ball, which means that some energy was lost during the collision.

Therefore, the fraction of the original kinetic energy transferred to the second ball is 0.75.

.280*v=.280v'+m*.140

solve for v'
Then, put that into...
1/2 .280 v^2=1/2 .280 v'^2 + 1/2 m .140^2
then solve for m.