An object traveling 200 feet per second slows to 50 feet per second in 5 seconds calculate the object
1 answer:
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Answer:twice of initial value
Explanation:
Given
spring compresses distance for some initial speed
Suppose v is the initial speed and k be the spring constant
Applying conservation of energy
kinetic energy converted into spring Elastic potential energy
When speed doubles
divide 1 and 2
Therefore spring compresses twice the initial value
The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.
The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.
The change in momentum of the system of the carts is 0.
<h3>Initial momentum of the carts before collision</h3>The momentum of the carts before the collision is calculated as follows;
P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s
P(blue) = 1.5 x 0 = 0
<h3>Momentum of the carts after collision</h3>The momentum of the carts after the collision is calculated as follows;
P(red) = 0.5 x 0.1 = 0.05 kgm/s
P(blue) = 1.5 0.1 = 0.15 kgm/s
<h3>Change in momentum of the carts</h3>ΔP = (0.05 + 0.15) - (0.2)
ΔP = 0
Learn more about momentum here: brainly.com/question/7538238