Inelastic Collision of 2 Carts - Displacement

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SUMMARY

The discussion centers on calculating the displacement of two colliding toy carts in an inelastic collision scenario. A 15.0 kg cart traveling at 5.5 m/s collides with a 10.0 kg cart moving at 4.0 m/s in the opposite direction, resulting in a combined velocity of 1.7 m/s post-collision. The friction coefficient is 0.40, leading to a deceleration of 3.92 m/s². The final calculation reveals that the carts slide 0.37 meters from the collision point before coming to a stop.

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Homework Statement


A 15.0 kg toy cart is traveling at 5.5 m/s to the right when it collides head on with a 10.0 kg cart traveling at 4.0 m/s to the left. The objects collide and stick togethere. The coefficient of friction between the wheels of the cart and the floor is 0.40. How far will the carts slide from the location of the collision?

ma = 15 kg
va = 5.5 m/s
mb = 10 kg
vb = 4.0 m/s
μ = 0.4

Homework Equations


Assuming right is positive.

mava + mbvb = ma+b * va+b
Fnet = ma
v2^2 = v1^2 + 2ad

The Attempt at a Solution


mava + mbvb = ma+b * va+b
15(5.5) + 10(-4) = 25 * va+b
42.5 = 25 * va+b
va+b = 1.7 m/s

Fnet = ma
Ff = ma
Fnμ = ma
- Since there is no vertical acceleration Fn = Fg
9.8(0.4) = a
a = -3.92 m/s

this is where I get stuck

Edit: I tried doing this but I'm not sure if I'm right.

v2^2 = v1^2 + 2ad
0 = (1.7)^2 + 2(-3.92)d
-2.89 = -7.84d
d = 0.37 m
 
Last edited:
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Yup. That works :smile:
 

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