A 3 kg cat sitting on a 1.5 kg piece of cardboard on a frozen lake wants to jump to shore without touching the ice. If there is no friction between the cardboard and the ice, when the cat jumps. the cardboard will move in the opposite direction with a velocity:

A. half as great as the cat's velocity. B. equal to the cat's velocity.
C. twice as great as the cat's velocity. D. four times as great as the cat's velocity

initial momentum = 0

so
final momentum = 0

0 = 3*Vcat + 1.5*Vcardboard
so
Vcardboard = 3 Vcat/1.5 = 2 Vcat

actually speed rather than velocity because directions are opposite.

To answer this question, we need to apply the principle of conservation of momentum. The momentum of an object is given by the product of its mass and velocity.

Let's analyze the situation step by step:

1. Initially, both the cat and the cardboard are at rest.
2. When the cat jumps, it exerts a force on the cardboard in one direction (let's call it the positive direction) due to Newton's third law of motion (for every action, there is an equal and opposite reaction).
3. By exerting this force, the cat imparts momentum to the cardboard in the positive direction, causing it to move.
4. According to the principle of conservation of momentum, the total momentum of the system (cat + cardboard) should be conserved. In other words, the momentum before the jump should be equal to the momentum after the jump.

Now, let's calculate the total momentum before and after the jump:

Before the jump:
- The cat's momentum is given by the product of its mass (3 kg) and velocity (0 m/s, since it is initially at rest): momentum(cat) = 3 kg × 0 m/s = 0 kg⋅m/s.
- The cardboard's momentum is given by the product of its mass (1.5 kg) and velocity (0 m/s, since it is initially at rest): momentum(cardboard) = 1.5 kg × 0 m/s = 0 kg⋅m/s.

After the jump:
- The cat's momentum will change due to its velocity after the jump (let's call it v_cat). So, the cat's momentum after the jump will be momentum(cat) = 3 kg × v_cat.
- The cardboard's momentum will also change due to its velocity after the jump (let's call it v_cardboard). So, the cardboard's momentum after the jump will be momentum(cardboard) = 1.5 kg × v_cardboard.

Since the total momentum before and after the jump should be the same, we can write an equation:

momentum(cat) + momentum(cardboard) = momentum(cat) + momentum(cardboard)
0 kg⋅m/s + 0 kg⋅m/s = 3 kg × v_cat + 1.5 kg × v_cardboard

Now, since the cat wants to jump to shore without touching the ice, we can assume that the cardboard moves in such a way that it cancels out the cat's momentum. Therefore, the velocity of the cardboard (v_cardboard) should be equal in magnitude (but opposite in direction) to the cat's velocity (v_cat). In other words:

v_cardboard = -v_cat

Now, let's substitute this relationship into the equation:

0 kg⋅m/s + 0 kg⋅m/s = 3 kg × v_cat + 1.5 kg × (-v_cat)

Simplifying the equation:

0 kg⋅m/s = 3 kg × v_cat - 1.5 kg × v_cat
0 kg⋅m/s = (3 kg - 1.5 kg) × v_cat
0 kg⋅m/s = 1.5 kg × v_cat

From this equation, we can see that the cat's velocity (v_cat) is zero, since multiplying it by any value will result in zero. Therefore, the cat cannot jump to shore without touching the ice.

To summarize, the correct answer to this question is: None of the given options (A, B, C, D) is correct. The cardboard will not move in the opposite direction because the cat's velocity is zero, and it cannot jump without touching the ice.