A 1.25 kg skateboard is coasting along the pavement at a speed of 4.24 m/s when a 1.00 kg cat drops from a tree vertically downward onto the skateboard. What is the speed of the skateboard-cat combination?
To find the speed of the skateboard-cat combination after the cat drops onto the skateboard, we can use the principle of conservation of momentum.
According to the law of conservation of momentum, the total momentum before the cat jumps onto the skateboard is equal to the total momentum after the cat lands on the skateboard.
The momentum of an object is defined as the product of its mass and velocity. Mathematically, momentum (p) can be calculated using the equation:
p = m * v
Where:
p = momentum
m = mass
v = velocity
Let's denote the mass of the skateboard as m1, the velocity of the skateboard as v1, and the mass of the cat as m2.
Before the cat lands on the skateboard, the momentum of the skateboard-cat system is given by the sum of the individual momenta:
p1 = m1 * v1
Once the cat lands on the skateboard, their combined momentum can be calculated as:
p2 = (m1 + m2) * v2
where v2 is the velocity of the skateboard-cat combination.
Since momentum is conserved, we can set the initial momentum equal to the final momentum:
p1 = p2
m1 * v1 = (m1 + m2) * v2
Now we can plug in the given values:
m1 = 1.25 kg (mass of the skateboard)
v1 = 4.24 m/s (velocity of the skateboard)
m2 = 1.00 kg (mass of the cat)
Substituting these values into the equation, we get:
(1.25 kg * 4.24 m/s) = (1.25 kg + 1.00 kg) * v2
Now we can solve for v2, the velocity of the skateboard-cat combination:
5.30 kg * m/s = 2.25 kg * v2
Divide both sides of the equation by 2.25 kg:
v2 = (5.30 kg * m/s) / 2.25 kg
v2 ≈ 2.36 m/s
Therefore, the speed of the skateboard-cat combination after the cat lands on the skateboard is approximately 2.36 m/s.