Here's a basketball problem: A 83.0 kg basketball player is running in the positive direction at 7.5 m/s. She is met head-on by a 101.6 kg player traveling at 5.3 m/s toward her. If the 101.6 kg player is knocked backwards at 3.4 m/s, what is the resulting velocity of the 83.0 kg player?

The speeds of the objects after elastic collision are

v₁= {-2m₂v₂₀ +(m₁-m₂)v₁₀}/(m₁+m₂)
v₂={ 2m₁v₁₀ - (m₂-m₁)v₂₀}/(m₁+m₂)

Well, it seems like these basketball players had a little run-in! Let's calculate the resulting velocity of the 83.0 kg player after this collision.

To start, let's use the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Before the collision:
Momentum of the 83.0 kg player = (83.0 kg) x (7.5 m/s)
Momentum of the 101.6 kg player = (101.6 kg) x (-5.3 m/s) (negative because he's heading backwards)

After the collision:
Momentum of the 83.0 kg player = (83.0 kg) x (vf) (let's call the resulting velocity vf)
Momentum of the 101.6 kg player = (101.6 kg) x (-3.4 m/s) (negative since he's still going backward)

Now, using the conservation of momentum, we can equate the two equations:

(83.0 kg) x (7.5 m/s) + (101.6 kg) x (-5.3 m/s) = (83.0 kg) x (vf) + (101.6 kg) x (-3.4 m/s)

Now let's do some math and solve for vf:

(623.5 kg·m/s) + (-538.48 kg·m/s) = 83.0 kg·vf + (-344.64 kg·m/s)

85.02 kg·m/s = 83.0 kg·vf + (-344.64 kg·m/s)

85.02 kg·m/s + 344.64 kg·m/s = 83.0 kg·vf

429.66 kg·m/s = 83.0 kg·vf

vf = 429.66 kg·m/s / 83.0 kg

vf ≈ 5.18 m/s

So, after this collision, the resulting velocity of the 83.0 kg player is approximately 5.18 m/s. Keep dribbling, my friend!

To solve this problem, we can use the law of conservation of momentum. This law states that the total momentum of an isolated system remains constant before and after a collision.

The momentum (p) of an object is defined as the product of its mass (m) and velocity (v):

p = m * v

Before the collision, the total momentum of the system is the sum of the individual momenta of the two players.

Let's calculate the initial momentum of the 83.0 kg player:

momentum_83kg_initial = 83.0 kg * 7.5 m/s = 622.5 kg·m/s

Similarly, let's calculate the initial momentum of the 101.6 kg player:

momentum_101.6kg_initial = 101.6 kg * 5.3 m/s = 538.48 kg·m/s

Since the collision is head-on, the 101.6 kg player gets knocked backwards at 3.4 m/s. This means the final velocity of the 101.6 kg player is -3.4 m/s (negative sign indicates direction).

Now, we can use the conservation of momentum to find the resulting velocity of the 83.0 kg player. The total initial momentum of the system should be equal to the total final momentum after the collision.

total_initial_momentum = momentum_83kg_initial + momentum_101.6kg_initial

After the collision, the 101.6 kg player has a momentum of:

momentum_101.6kg_final = 101.6 kg * (-3.4 m/s) = -345.44 kg·m/s

The final momentum of the 83.0 kg player can be calculated as:

momentum_83kg_final = total_initial_momentum - momentum_101.6kg_final

Now, we can find the final velocity of the 83.0 kg player by dividing the final momentum by the player's mass:

v_final_83kg = momentum_83kg_final / 83.0 kg

Calculating the values:

total_initial_momentum = 622.5 kg·m/s + 538.48 kg·m/s
total_initial_momentum = 1160.98 kg·m/s

momentum_83kg_final = 1160.98 kg·m/s - (-345.44 kg·m/s)
momentum_83kg_final = 1506.42 kg·m/s

v_final_83kg = 1506.42 kg·m/s / 83.0 kg
v_final_83kg ≈ 18.14 m/s

Therefore, the resulting velocity of the 83.0 kg player is approximately 18.14 m/s.

To solve this problem, we can use the principle of conservation of momentum. The principle states that the total momentum before a collision is equal to the total momentum after the collision, assuming no external forces act on the system.

Momentum, denoted by 'p', is calculated by multiplying the mass of an object by its velocity. We can express this mathematically as:

p = m * v

Considering the collision of the basketball players, we need to calculate the momentum of each player before and after the collision. Let's define the positive direction as the direction of the 83.0 kg player's initial velocity.

The momentum of the 83.0 kg player before the collision is given by:

p1_before = m1 * v1_before

where m1 is the mass of the 83.0 kg player and v1_before is her initial velocity.

Similarly, the momentum of the 101.6 kg player before the collision is given by:

p2_before = m2 * v2_before

where m2 is the mass of the 101.6 kg player and v2_before is his initial velocity.

Since the players are moving towards each other, the momentum of the 101.6 kg player could be considered negative:

p2_before = -m2 * v2_before

After the collision, the 101.6 kg player is knocked backward at 3.4 m/s. We can denote this velocity as v2_after. The resulting velocity of the 83.0 kg player after the collision can be denoted as v1_after.

Since momentum is conserved, we have:

p1_before + p2_before = p1_after + p2_after

Combining the equations, we have:

m1 * v1_before + (-m2 * v2_before) = m1 * v1_after + (-m2 * v2_after)

Plugging in the given values, we have:

83.0 kg * 7.5 m/s + (-101.6 kg * 5.3 m/s) = 83.0 kg * v1_after + (-101.6 kg * 3.4 m/s)

Now, we can rearrange the equation to solve for v1_after, the resulting velocity of the 83.0 kg player:

83.0 kg * v1_after = (83.0 kg * 7.5 m/s + (-101.6 kg * 5.3 m/s)) + (101.6 kg * 3.4 m/s)

Now, let's calculate the resulting velocity.

83.0 kg * v1_after = (622.5 kg•m/s - 539.68 kg•m/s) + (344.64 kg•m/s)

83.0 kg * v1_after = 427.82 kg•m/s

v1_after = 427.82 kg•m/s / 83.0 kg

v1_after ≈ 5.15 m/s

So, the resulting velocity of the 83.0 kg player after the collision is approximately 5.15 m/s in the positive direction.