Glider A of mass 0.355 kg moves along a frictionless air track with a velocity of 0.095 m/s.
1 answer:
Answer:
vB' = 0.075[m/s]
Explanation:
We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.
Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.
where:
mA = 0.355 [kg]
vA = 0.095 [m/s] before the collision
mB = 0.710 [kg]
vB = 0.045 [m/s] before the collision
vA' = 0.035 [m/s] after the collision
vB' [m/s] after the collison.
The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.
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The speed is 0.956 m / s.
<u>Explanation</u>:
The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.
K.E = 1/2 m
where K.E represents the kinetic energy,
m represents the mass,
v represents the velocity.
K.E = 1/2 m
1.10 10^42 = 1/2
3.26
10^31
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2) / (3.26
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v = 0.956 m / s.