Cart 1 with mass 4.44 kgcollides with a stationary cart (cart 2) of mass 2.81 kg

Question

Cart 1 with mass 4.44 kgcollides with a stationary cart (cart 2) of mass 2.81 kg. Beforethe

collision cart 1 has avelocity of 4.94 m/s to the right. After the collision, cart 1 hasa velocity

1.11 m/s to the right andcart 2 has a velocity 6.05 m/s to the right.

a. What is the totalmomentum of the two cart system before the collision?

b. What is the totalkinetic energy of the two cart system before thecollision?

c. What is the totalmomentum of the two cart system after the collision?

d. What is the totalkinetic energy of the two cart system after thecollision?

Explanation / Answer

a. So total momentum before = p = mv(first car) + mv(secondcar) = 4.44*4.94 + 2.81*0 = 21.93 kgm/s b. Since second cart is stationary, the only kinetic energy isfound in the first cart: KE = 1/2mv^2 = 1/2(4.44)(4.94)^2 = 54.18J c. Total momentum after = p = mv(firstcar) + mv(secondcar) =4.44(1.11) + 2.81(6.05) = 21.93 kgm/s d. Total KE = KEfirst + KEsecond = 1/2mv^2(first) +1/2mv^2(second) = 1/2(4.44)(1.11)^2 + 1/2(2.81)(6.05)^2 =54.16 e. I would say this collision is elastic because momentum andkinetic energy were conserved. This is the major difference betweenelastic and inelastic collisions. In inelastic collisions, energyis lost to other forms (heat, light, sound), and since there was noloss in energy in this collision (the difference between the twoKEs are negligible, possibly a data collection error), thiscollision is elastic.
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