Show all work to receive credit. Scrap paper will be provided. Two air carts hav
ID: 1836523 • Letter: S
Question
Show all work to receive credit. Scrap paper will be provided. Two air carts have masses m_1 = 127 grams and m_2 = 205 grams (the mass of any additional weights is included in m_1 and m_2). Assume that there are absolutely no frictional or other external, horizontally directed forces acting on the system of the two air carts. Initially, car #1 is moving to the right and cart 2 is moving to the left. Cart #1, which has a mounted flag 14.00 cm long, takes 0.283 seconds to go through its photogate, and Cart #2, with an identical flag on top, takes 0.333 seconds to go through its photogate. After the collision the two carts stick together, i.e., they make completely inelastic collision. What is the velocity of each cart before the collision Please report both the magnitude and direction of the velocity of each cart and remember to include units! Make sure you designate which direction is positive. Using your knowledge of physics, determine the expected total momentum of the system after the collision. Remember units? Determine the velocity of the two cars after the collision (i.e., their final velocity). Remember units! Determine the expected total kinetic energy of the system after the collision. Make sure you designate which direction is positive. Remember units!Explanation / Answer
a) Considering right side as +ve direction then .
Speed of cart1, u1 = (14 x 10^-2 m ) / (0.283 sec)
u1 = 0.495 m/s to the right
so velcity = + 0.495 m/s
for cart2:
u2 = - 0.14 / 0.333 = - 0.42 m/s
(minus indicates that direction is to the left)
b) before collision,
total momentum = m1u1 + m2u2
= (0.127 x 0.495) + (0.205 x -0.42)
= - 0.0232 kg m/s
and during collision, momentum of system will not change .
hence momentum of system just after collision = - 0.0232 kg m/s
c)
after collison both moves with v then
momentum = (0.127 + 0.205)v = - 0.0232
v = - 0.07 m/s
(minus sign indicates left direction)
d) KE after collision = (0.127 + 0.205)(0.07^2)/2
= 8.13 x 10^-4 J