A 1.22 kg skateboard is coasting along the pavement at a speed of 5.84 m/s when a 1.20 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.
The principle of conservation of momentum states that the total momentum before an event is equal to the total momentum after the event, as long as no external forces are acting on the system. Momentum is a vector quantity and is defined as the product of an object's mass and its velocity.
Therefore, we need to calculate the initial momentum of the skateboard and the final momentum of the skateboard-cat combination, and equate them to find the final speed.
1. Calculate the initial momentum of the skateboard:
Momentum (Skateboard) = mass (Skateboard) x velocity (Skateboard)
= 1.22 kg x 5.84 m/s
2. Calculate the final momentum of the skateboard-cat combination:
Since the cat drops vertically downward onto the skateboard, we can assume that the cat and skateboard move together as one system. The final momentum is the sum of the momentum of the cat and the momentum of the skateboard.
Momentum (Skateboard-Cat) = Momentum (Skateboard) + Momentum (Cat)
= (mass (Skateboard) x velocity (Skateboard)) + (mass (Cat) x velocity (Cat))
= (1.22 kg x 5.84 m/s) + (1.20 kg x 0 m/s)
3. Simplify and solve for the final velocity:
Momentum (Skateboard-Cat) = (1.22 kg x 5.84 m/s) + (1.20 kg x 0 m/s)
= 7.1248 kg·m/s
Now, we need to divide the total momentum by the total mass to find the final velocity of the skateboard-cat combination:
Final Velocity (Skateboard-Cat) = Momentum (Skateboard-Cat) / Total Mass (Skateboard + Cat)
= 7.1248 kg·m/s / (1.22 kg + 1.20 kg)
Simplifying that equation will give you the final speed of the skateboard-cat combination.
To determine the speed of the skateboard-cat combination after the cat drops onto it, we can use the principle of conservation of momentum.
The momentum before the cat drops onto the skateboard is equal to the momentum after the drop.
The momentum (p) of an object is given by the formula: p = m * v, where m is the mass of the object and v is its velocity.
Let's denote the initial velocity of the skateboard as v1, the mass of the skateboard as m1, and the mass of the cat as m2.
Given:
Mass of the skateboard (m1) = 1.22 kg
Velocity of the skateboard (v1) = 5.84 m/s
Mass of the cat (m2) = 1.20 kg
Using the conservation of momentum equation, we can write:
(m1 * v1) = (m1 * v1') + (m2 * v2'),
where v1' is the final velocity of the skateboard-cat combination, and v2' is the final velocity of the cat.
Rearranging the equation, we have:
m1 * v1 - m2 * v2' = m1 * v1'
Plugging in the given values, we get:
(1.22 kg * 5.84 m/s) - (1.20 kg * v2') = (1.22 kg + 1.20 kg) * v1'
Simplifying:
(7.1328 kg*m/s) - (1.20 kg * v2') = (2.42 kg) * v1'
Now, we can solve for v1':
7.1328 kg*m/s - 1.20 kg * v2' = 2.42 kg * v1'
Dividing both sides of the equation by 2.42 kg, we get:
2.945 m/s - 0.495 * v2' = v1'
Therefore, the final velocity of the skateboard-cat combination (v1') is equal to 2.945 m/s minus 0.495 times the final velocity of the cat (v2').