A 160 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm and released. A student with a stopwatch finds
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Answer:
20 hertz of frequency produced.
Explanation:
Here we will find frequency and period should be in second, here given: 0.05 seconds
using the formula:
According to the law of conservation of momentum:
m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision
Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.
m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?
Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.
The velocity of the 2nd car after the collision is 0.03m/s.
m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision
Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.
m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?
Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.
The velocity of the 2nd car after the collision is 0.03m/s.