Two carts on a track have the same mass m = 0.90 kg. They collide completely ine
ID: 1974861 • Letter: T
Question
Two carts on a track have the same mass m = 0.90 kg. They collide completely inelastically with Cart 1 initially traveling at v1i = 0.93 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 = 0.15 s.The Momentum of the cart before, p initial=mv initial, and after, p final=mv final, the collision.
p initial = _____________ ± ________________
p final= _______________ ± ________________
The Change in momentum ?p = pf – pi of the cart.
?p = __________ ± _____________
If the collision lasts a time interval ?t = 0.110±0.007 s, calculate the average force F ave = ?p/?t.
F ave= ____________ ± _______________
Explanation / Answer
completely inelastic means they stick together and move as one object the kinetic energy will obviously not be conserved but the momentum will be conserved whenever there is no extermal force initial momentum = 0.9 * 1.51 final momentum = (.9+.9) (v)= .18v so .18v = .9*1.51 implies v = .755 m/s 1. What is the final velocity of Cart 1? v1f = .755 m/s 2. What is the final velocity of Cart 2? v2f =.755 m/s since both move together as single object wiuth same speed what is the initial kinetic energy of cart 1 = 1/2(.9)(1.51)^2 = 1.026045 J initial kinetic energy of cart 2 = 0 J final kinetic energy of cart 1 = 1/2 (.9) (.755)^2 = 0.25651125 Joules final kinetic energy of cart 2 = same = 0.25651125 J change in kinetic energy of cart 1 = -0.76953375 J change in kinetic energy of cart2 = .25651125 J 3. What is the change in kinetic energy of each Cart? ?K1 =-0.76953375 J ?K2 =.25651125 J Impulse is change in momentum F.(t) = change in momentum 4. What is the impulse delivered to cart 2? ?p2 = (.9)(.755) - 0 = .6795 5. What is the average force of Cart 1 on Cart 2 during the collision? Average F12 = change in momentum / time of contact = .6795/.05 = 13.59 Newton 6. What is the average force of Cart 2 on Cart 1 during the collision? Average F21 = 13.59 Newton you can either calculate the change in momentm of cart 1 and divide it by the time of contact, or according to newton's third law the force exerted by cart 2 on cart 1 is equal and opposite to force exerted by cart 1 on cart 2 all the best
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