A student is shopping for study snacks at a local grocery store. Somebody broke a very large bottle of olive oil on the floor. The student slips on the oil, sliding along the aisle. A helpful clerk rushes to the rescue, but also slips and crashes into the student. Entangled, they slide along the floor together. Assume no friction or air drag.
If before the collision the student (weighing 41.2 kg) (indicated orange in the figure above) was moving at 1.68 m/s, and the clerk (weighing 45.3 kg) (indicated green) was moving at 3.26 m/s, how fast will they be moving together after the collision?
- AshLv 72 months agoFavourite answer
Conservation of momentum in x direction
m₁ux₁ + m₂ux₂ = (m₁+m₂)vx
vx = (m₁ux₁ + m₂ux₂)/(m₁+m₂)
vx = [(41.2 kg)(1.68 m/s) + (45.3 kg)(0)] / (41.2 kg + 45.3 kg)
vx = 0.800 m/s
Conservation of momentum in y direction
m₁uy₁ + m₂uy₂ = (m₁+m₂)vy
vy = (m₁uy₁ + m₂uy₂)/(m₁+m₂)
vy = [(41.2 kg)(0) + (45.3 kg)(3.26 m/s)] / (41.2 kg + 45.3 kg)
vy = 1.71 m/s
Resultant velocity = √[(0.800)² + (1.71)²] = 1.89 m/s ← They are moving at this velocity
- ?Lv 72 months ago
Conservation of momentum shall apply :
Final momentum M :
M = √ (41.2*1.68)^2+(45.3*3.26)^2 = 163 kg*m/sec
Final velocity V = M/(WF1+WF2) = 163/(41.2+45.3) = 1.88 m/sec
- JimLv 72 months ago
A picture is worth...
Remember to do the vector addition.