A 1.40 kg skateboard is coasting along the pavement at a speed of 5.5 m/s when a 0.800 kg cat drops from a tree vertically downward onto the skateboard. What is the speed of the skateboard-cat combination.
momentum is conserved.
in the direction of movement
1.4*5.5+.8*0= (1.4+.8)V
solve for V
3.26
To find the speed of the skateboard-cat combination after the cat drops onto the skateboard, we'll apply the law of conservation of momentum.
The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it.
In this case, the initial momentum of the skateboard is equal to the final momentum of the skateboard-cat combination.
The momentum (p) is calculated as the product of an object's mass (m) and its velocity (v):
p = m * v
Let's denote the initial velocity of the skateboard as v1 and the final velocity of the skateboard-cat combination as v2.
The initial momentum of the skateboard is given by:
p1 = m1 * v1
The final momentum of the skateboard-cat combination is:
p2 = (m1 + m2) * v2, where m1 is the mass of the skateboard (1.40 kg) and m2 is the mass of the cat (0.800 kg).
Using the law of conservation of momentum, we can set the initial momentum equal to the final momentum:
m1 * v1 = (m1 + m2) * v2
Now, we can solve for v2, which represents the speed of the skateboard-cat combination.
v2 = (m1 * v1) / (m1 + m2)
Plugging in the given values:
v2 = (1.40 kg * 5.5 m/s) / (1.40 kg + 0.800 kg)
Calculating this expression will give us the speed of the skateboard-cat combination.
To find the speed of the skateboard-cat combination, we can use the principle of conservation of momentum. The total momentum before the cat drops onto the skateboard is equal to the total momentum after the cat drops onto the skateboard.
The momentum of an object is calculated by multiplying its mass by its velocity.
Let's denote the initial velocity of the skateboard as v1, the mass of the skateboard as m1, the final velocity of the skateboard-cat combination as v2, and the mass of the cat as m2.
According to the conservation of momentum, the initial momentum is equal to the final momentum:
(m1 * v1) + (m2 * 0) = (m1 * v2) + (m2 * v2)
Since the cat drops vertically downward, its initial velocity is 0 m/s. Therefore, the equation becomes:
(m1 * v1) = (m1 * v2) + (m2 * v2)
Now, we can substitute the values given in the problem:
(1.40 kg * 5.5 m/s) = (1.40 kg * v2) + (0.800 kg * v2)
Simplifying the equation, we get:
7.70 kg m/s = (2.20 kg * v2)
Dividing both sides of the equation by 2.20 kg, we find:
v2 = (7.70 kg m/s) / 2.20 kg
v2 ≈ 3.50 m/s
Therefore, the speed of the skateboard-cat combination is approximately 3.50 m/s.