3. A 0.5 kg cart is launced with a 1-1 . velocity of 3 m\'s toward a stationary

Question

3. A 0.5 kg cart is launced with a 1-1 . velocity of 3 m's toward a stationary 1.2 kg cart. This is done three tmes: -he first time the carts stick together .he second time the 0.5 kg cart bounces backward at a speed of 2 m/s; For each case, draw a table analyzing these collisions in terms of the velocity and momertu of each cart before and after the collisions. Your table should include units and clearly label which cart is which, and which valucs represent thc "teforc" and which ones represent the "after."

Explanation / Answer

given, mass of cart m = 0.5 kg
initial velocity v = 3 m/s
mass of stationary cart, m2 = 1. 2 kg

case 1 : the carts stick together
befopre the collision
momentum of cart1 = m*v = 1.5 kg m/s
momentum of cart 2 = m2*0 = 0 kg m/s
after collision let the speed of both cards be u
then momentum of both cars = (m + m2)u
now from conservation of momentum
(0.5 + 1.2)u = 1.5
u = 0.8823 m/s
hence
final momentum of two cars - 0.8823 kg m/s
final speed opf both carts, u = 0.8823 m/s

case 2: cart 1 bounces back with speed v' = 2 m/s
befopre the collision
momentum of cart1 = m*v = 1.5 kg m/s
momentum of cart 2 = m2*0 = 0 kg m/s
after the collision let the speed of cart 2 be u
then from conservation of momentum
1.5 = -mv' + m'*u
1.5 + 0.5*2 = 1.2*u
u = 2.0833 m/s
hence final speed of cart 2 = 2.0833 m/s
final momentum of cart 1 = -0.5*2 = -1 kg m/s
final momentu mof cart 2 = 2.5 kg m/s

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.