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# An 8kg kart is traveling at 2 m/s east and collides with a 4kg cart traveling 5 m/s west.?

An 8kg kart is traveling at 2 m/s east and collides with a 4kg cart traveling 5 m/s west. The 8kg cart locks together with the 4 kg cart after the collision.

a. Find the final velocity of the combined carts.

b. Determine the force of the 8kg cart is the collision occurs in .1 seconds.

### 2 Answers

- billrussell42Lv 73 years agoFavorite Answer
conservation of momentum

taking E to be positive...

initial momentum = 8•2 – 4•5 = 16–20 = –4 kgm/s

final momentum is the same

V(8+4) = –4

V = –4/12 = –1/3 m/s, pointed west

change in momentum is impulse

But this is not "force of the 8kg cart" as carts do not have forces. This is the force between the two during the collision.

I think you want the force ON the 8 kg cart as it undergoes the collision and change in momentum.

initial momentum is 16 kgm/s

final momentum is 8•(–1/3) = –8/3 kgm/s

change in momentum is 16 + 8/3 = 18.67 kgm/s

Force is change in momentum / time = 18.67 / 0.1 = 186.7 N

- az_lenderLv 73 years ago
(a) [(4 kg)(5 m/s) - (8 kg)(2 m/s)]/(12 kg) = 0.33 m/s westward.

(b) (4 kg)(5 m/s - 0.33 m/s)/(0.1 s) = about 190N but use a calculator.