A 10.0 kg and a 2.0 kg cart approach each other on a horizontal frictionless air track. Their total kinetic energy before collision is 96 J. Assume their collision is elastic. What is the final speed in m/s of the 10.0 kg mass if that of the 2.0 kg mass is 8.0 m/s? (Hint: There are 2 conditions for elastic collisions.)

2.53 m/s

Explanation:

From the law of conservation of momentum,

For an Elastic collision,

Total kinetic energy before collision = Total kinetic energy after collision

Ek₁ = 1/2mv²+1/2m'v'² Equation 1

Where Ek₁ = total kinetic energy before collision, m = mass of the first cart, v = final velocity of the first cart, m' = mass of the second cart, v' = final velocity of the second cart.

Given: Ek₁ = 96 J, m = 10 kg, m' = 2 kg, v' = 8 m/s.

Substitute into equation 1 and solve for the value of v.

96 = 1/2(10)(v²)+1/2(2)(8²)

96 = 5v²+64

5v² = 96-64

5v² = 32

v² = 32/5

v² = 6.4

v = √6.4

v = 2.53 m/s

Hence the final speed of the 10 kg mass = 2.53 m/s

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation