Visualize a frozen ice rink under a faintly coloured twilight sky. Two figure skaters, a somewhat heavier Middle-Eastern man in vibrant winter attire and a lighter South Asian woman in an elegant ice-skating outfit, are positioned facing each other holding hands. Their facial expressions reveal concentration and determination. Beneath them, the ice sparkles under the light, alluding to its frictionless nature. As they prepare to push off one another, highlight the sense of kinetic energy about to be unleashed. Ensure the surroundings such as snow-dusted trees and a rustic wooden viewing gallery with spectators bundled up in warm clothes.

Two ice skaters want to prove conservation of momentum. Skater A has a mass of 72 kg, and skater B has a mass of 48 kg. They are originally at rest. They hold hands and face each other. They then push off each other and move in opposite directions. Skater B moves with a velocity of 3 m/s.

What must the velocity of skater A be in the opposite direction after the push in order to prove conservation of momentum?

(1 point)

4.5 m/s

3.0 m/s

1.5 m/s

2.0 m/s

6 answers

It's 2.0m/s

So no answer?

Hmm, let me calculate this with a touch of humor.

Since we're dealing with conservation of momentum, it's like a delicate ice dance between the two skaters. Skater A, weighing in at a hefty 72 kg, needs to show some serious moves. Skater B, at a svelte 48 kg, seems to prefer a leisurely stroll on ice.

Now, when they push off each other, they will go in opposite directions. To maintain the momentum balance, the product of their masses and velocities must be equal on both sides. In this case, Skater A's mass (72 kg) must be multiplied by an unknown velocity to balance out Skater B's mass (48 kg) multiplied by his velocity (-3 m/s because he is moving in the opposite direction).

Let's work it out!

(72 kg) x (V_A) = (48 kg) x (-3 m/s)

Now let's solve for V_A, which represents the velocity of Skater A in the opposite direction after the push.

V_A = (48 kg x -3 m/s) / 72 kg

Calculating that gives us V_A ≈ -2.0 m/s.

So, the answer is 2.0 m/s, which means good old Skater A is gliding in the opposite direction with a velocity of 2.0 m/s. Keep on skating, Skater A, and keep that momentum conserved!

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before an event is equal to the total momentum after the event.

In the initial state, both skaters are at rest, so their total momentum is zero.

After the push, Skater A and Skater B move in opposite directions. Let's denote the unknown velocity of Skater A as V_A.

The momentum of Skater A after the push is given by mass of Skater A (m_A) times velocity of Skater A (V_A).
So, the momentum of Skater A after the push is given by: P_A = m_A * V_A.

The momentum of Skater B after the push is given by mass of Skater B (m_B) times velocity of Skater B (V_B).
So, the momentum of Skater B after the push is given by: P_B = m_B * V_B.

According to the principle of conservation of momentum, the sum of the momenta before the push should be equal to the sum of the momenta after the push.
So, we can write the equation:

0 = P_A + P_B
0 = m_A * V_A + m_B * V_B

Substituting the given values, we have:

0 = 72 kg * V_A + 48 kg * 3 m/s

Simplifying the equation:

0 = 72 kg * V_A + 144 kg * m/s

Rearranging the equation to solve for V_A:

72 kg * V_A = -144 kg * m/s

V_A = (-144 kg * m/s) / 72 kg

V_A = -2 m/s

Therefore, the velocity of Skater A in the opposite direction after the push is 2.0 m/s.

The correct answer is: 2.0 m/s.