Two carts on a track have the same mass m = 2.20 kg . They collide completely in

Question

Two carts on a track have the same mass m = 2.20 kg. They collide completely inelastically with Cart 1 initially traveling at  v1i = 0.56 m/s to the right and Cart 2 initially at rest (v2i = 0 m/s). The collision is monitored by two motion sensors. The time over which the collision lasts is t = 0.11 s.

What is the final velocity of Cart 1?

v1f =

What is the final velocity of Cart 2?

v2f =

What is the change in kinetic energy of each Cart?

K1 =   
K2 =   

What is the impulse delivered to cart 2?

p2 =

What is the average force of Cart 1 on Cart 2 during the collision?

Average F12 =

What is the average force of Cart 2 on Cart 1 during the collision?

Average F21 =

Explanation / Answer

1)

since the collision is enelastic, the two carts will stick togeather
Thus by the law of momentum conservation:-
m1u1 + m2u2 = (m1+m2) x v
=>2.20 x 0.56 + 0 = 2 x 2.20 x v
=> v = 0.28 m/s

2) same as of cart1 = 0.28 m/s

3)Final KE = 1/2m(v2^2) - 1/2m(v2^2)

K1 = 1/2 x 2.20 x 0.28 x 0.28 - 1/2 x 2.20 x 0.56 x 0.56 =-0.2587 J[-ve=>decreased]

K2 = 1/2 x 2.20 x 0.28 x 0.28 - 0 = 0.08624J [+ve=>increased]

4) Impulse (p) = m x v
=> p2 = 2.20 x (0.28 - 0) = 0.616 N-s

5)Impulse/t = Force
=>F12 = 0.616/0.11 = 5.6 N

6) By F12 = -F21 [Newtons 3rd law]
=>F21 = -5.6 N [-ve=>opposite direction to F12]

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