A small, 300 g cart is moving at 1.10 m/s on an air track when it collides with a larger, 4.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision (answer in m/s please)?

According to the question:

Mass,

Final velocity,

Initial velocity,

By using the law of conservation, we get

→

By substituting the values, we get

→

→              

Thus the response above is appropriate.

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0.0158m/s

Explanation:

Using the law of conservation of energy which states that the sum of momentum before collision is equal to the sum after collision. It is expressed mathematically as;

m1u1 + m2u2 =  m1v1 + m2v1

m1 and m2 are the masses of the object

u1 and u2 are the initial velocities

v1 and v2 are the final velocities

Given

m1 = 300g  = 0.3kg

u1 = 1.10m/s

m2 = 4.00kg

u2 = 0m/s (at rest)

v1 = 0.890

v2 = ?

Substitute the given values into the formula;

0.3(1.10) + 0 = 0.3(0.89) + 4v2

0.33 = 0.267 + 4v2

0.33-0.267 = 4v2

0.063 = 4v2

v2 = 0.063/4

v2 = 0.0158m/s

Hence the speed of the large cart after the collision is 0.0158m/s

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2