4. A 5.0 kg block moving to the right at 12.0 m/sec collides with a 4.0 kg block moving to the right at 2 m/sec. If the 4.0 kg moves to the right at 10 m/sec after the collision, what will be the speed of the 5.0 kg block? How much energy was lost during the collision?

Energy lost = 55.28Joules

Explanation:

Using the law of momentum:

m1u1+m2u2 = m1v1+m2v2

m1 and m2 are the masses of the objects

u1 and u2 are the initial velocities.

v1 and v2 are final velocities

Given

m1 = 5kg

m2 = 4kg

u1 = 12m/s

u2 = 2m/s

v1 = 10m/s

v2 = ?

Substitute

5(12)+4(2)=4(10)+5v2

60+8 = 40+5v2

68-40 = 5v2

28 = 5v2

v2 = 28/5

v2 = 5.6m/sec

Hence the speed of the 5kg block is 5.6m/s

Energy lost = Kinetic energy after collision - kinetic energy before collision

KE after collision = 1/2m1v1²+1/2m2v2²

KE after collision = 1/2(5)10²+1/2(4)(5.6)²

= 250+62.72

= 312.72Joules

KE before collision = 1/2m1u1²+1/2m2u2²

KE before collision = 1/2(5)12²+1/2(4)(2)²

= 360+8

= 368Joules

Energy lost = 368-312.72

Energy lost = 55.28Joules

this could be newton#'s law of cooling. the cooling curve is exponential

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