Question

Two penguins are having fun sliding on an icy, nearly frictionless slope, that rises above level ground at an angle of 11 degrees. One penguin has mass 15 kg while the other has mass 5 kg. At one point, the heavier penguin starting from rest, begins sliding frictionlessly down the slope. At the same time, the lighter penguin, starting from higher up the slope launches itself downward. It catches up to the heavier penguin after 2 seconds, both penguins then collide and, slide together down the hill. Immediately after the collision the pair of penguins is moving down the slope with a speed of 12 m/s. What is the total momentum of the two penguins immediately after their collision? What is the heavier penguin's momentum just before the collision? What was the lighter penguin's momentum just before colliding with the heavier one? How fast did the lighter penguin launch itself down the slope? How much total impulse was exerted on the two penguins by gravity before the collision?

Answer 1

m₁ = mass of heavy penguin = 15 kg

m₂ = mass of light penguin = 5 kg

V = velocity of penguins after collision =12 m/s

Total momentum after collision is given as

P_f = (m₁ + m₂) V = (15 + 5) (12) = 240 kgm/s

for the heavy penguin :

a = acceleration = g Sin11 = (9.8 Sin11)= 1.87 m/s²

v_o1 = initial velocity of heavy penguin = 0 m/s

t = 2 sec

velocity just before collision is given as

v₁ = v_o1 + at = 0 + (1.87) (2) = 3.74 m/s

v₂ = velocity of lighter penguin just before collision = ?

using conservation of momentum

m₁ v₁ + m₂ v₂ = (m₁ + m₂) V

(15)(3.74) + (10) v₂ = 240

v₂ = 18.4 m/s

v_o2 = launch velocity of lighter penguin

using the equation

v₂ = v_o2 + a t

18.4 = v_o2 + (1.87) (2)

v_o2 = 14.7 m/s