A1000 kilogram empty cart moving with a speed of 6.0 meters per second is about to collide with a stationary loaded cart having a total mass of 5000. kilograms. after the collision, the carts lock and move together. [assume friction is negligible.]
a) calculate the speed of the combined carts after the collision. [show all work, including the equation and substitution with units.]
b) calculate the kinetic energy of the combined carts after the collision. [show all work, including the equation and substitution with units.]
c) how does the kinetic energy of the combined carts after the collision compare to the kinetic energy of the carts before the collision?

1) v = 1m/s

2) K.E = 3 kJ

3)

Explanation:

From the law of conservation of momentum, p = mv

1) For the given question, ------- eqn 1

substitute mv for p in eqn 1,

and

To solve for p, which is the final momentum after collision

------ eqn 2

Parameters

= 1000 kg

= 5000 kg

= 6 m/s

= 0 (static object)

substitute for all m and v in eqn 2

p =

p = 6000 + 0

p = 6000 kgm/s

Recall p = mv

where p = 6000kgm/s and m =

v = ?

6000 kgm/s = (1000 + 5000)v

6000 = 6000v

v = 1m/s

2) K.E =

Therefore, K.E after collision

K.E =

K.E = 0.5 X 6000 X 1 = 3000 J

K.E = 3 kJ

3) Comparing K.Es of the carts, first we write out the K.E of each cart and their combined K.E

(Please ignore the Armstrong in the following equation, I don't know how it appeared from the equation tool)

= 18 kJ

= 0

Combined K.E = 3 kJ

From the above K.Es,

i) The K.E of the combined cart after collision is lesser than the K.E of the first cart of 1000 kg.

3 kJ < 18 kJ

ii) The K.E of the combined cart is greater than the K.E of the second cart of 5000 kg.

0 kJ < 3 kJ

Conclusion;

i.e 0 kJ < 3 kJ < 18 kJ

total distance/total time

(85 + 82)/(5+3)

167/8

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