A 8.0 kg cart was observed moving north with a speed of 12.2 m/s before it colli

Question

A 8.0 kg cart was observed moving north with a speed of 12.2 m/s before it collided with an identical cart moving south at 8.1 m/s. The collision was observed to cause the carts to couple and move together as one.

a) What is in kg m/s the combined momentum of both carts before the collision?

b) What is in kg m/s the combined momentum of both carts after the collision?

c) What is the speed in m/s of the carts after the collision?

d) What is in Joules the kinetic energy of the system of both carts before the collision?

e) What is in Joules the kinetic energy of the system of both carts after the collision?

Explanation / Answer

a)

the combined momentum of both carts before the collision = 8*12.2 + 8*8.1 = 162.4 kg m/s

b)

the combined momentum of both carts before the collision = 162.4 kg m/s because momentum is conserved.

C)

8*12.2 + 8*8.1 = (8+8)*v

=> v = 162.4/16 = 10.15 m/s

d)

the kinetic energy of the system = 1/2*8*12.22 + 1/2*8*8.12 = 857.8 J

e)

the kinetic energy of the system of both carts after the collision = 1/2*16*10.152 = 824.18

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