A 1.25 kg skateboard is coasting along the

pavement at a speed of 5.04 m/s when a 0.640
kg cat drops from a tree vertically downward
onto the skateboard.
What is the speed of the skateboard-cat
combination?
Answer in units of m/s Your answer must
be within ± 1.0%

conserve momentum:

1.25*5.04 = (1.25+0.640)v

To find the speed of the skateboard-cat combination, we can use the principle of conservation of momentum.

The initial momentum of the skateboard is given by:
p1 = m1 * v1
where m1 is the mass of the skateboard (1.25 kg) and v1 is its initial velocity (5.04 m/s).

The momentum of the cat just before it lands on the skateboard is given by:
p2 = m2 * v2
where m2 is the mass of the cat (0.640 kg) and v2 is its velocity just before landing.

Since momentum is conserved, the total initial momentum of the system (skateboard + cat) should be equal to the total final momentum. Therefore, we have:
p1 + 0 = (m1 + m2) * v_final

Rearranging the equation, we can solve for the velocity of the skateboard-cat combination (v_final):
v_final = (p1 + 0) / (m1 + m2)
v_final = p1 / (m1 + m2)

Substituting the given values, we get:
v_final = (1.25 kg * 5.04 m/s) / (1.25 kg + 0.640 kg)

Evaluating the expression, we can find the speed of the skateboard-cat combination.

To find the speed of the skateboard-cat combination, we can first calculate the momentum of the skateboard and the cat separately before the collision, and then calculate the momentum of the combined system after the collision.

The momentum of an object can be calculated by multiplying its mass by its velocity.

Given:
Mass of skateboard (m1) = 1.25 kg
Velocity of skateboard (v1) = 5.04 m/s
Mass of cat (m2) = 0.640 kg

The momentum of the skateboard before the collision (p1) is:
p1 = m1 * v1

The momentum of the cat before the collision (p2) is:
p2 = m2 * 0 (as the cat is initially at rest)

As the cat drops vertically downward onto the skateboard, the total momentum of the system is conserved. Therefore, the momentum of the combined system after the collision (p3) is equal to the sum of the momenta of the skateboard and the cat before the collision.

p3 = p1 + p2

Now, let's calculate the momenta:

p1 = (1.25 kg) * (5.04 m/s) = 6.30 kg·m/s
p2 = (0.640 kg) * 0 = 0 kg·m/s

p3 = 6.30 kg·m/s + 0 kg·m/s = 6.30 kg·m/s

Next, to find the speed of the skateboard-cat combination, we divide the total momentum by the combined mass of the system.

Mass of the combined system = mass of skateboard + mass of cat
m3 = m1 + m2

m3 = 1.25 kg + 0.640 kg = 1.890 kg

Now, we can calculate the speed of the combined system (v3):

v3 = p3 / m3

v3 = 6.30 kg·m/s / 1.890 kg

v3 ≈ 3.33 m/s

Therefore, the speed of the skateboard-cat combination is approximately 3.33 m/s.

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