An 80 kg hunter gets a rope around a 400 kg
polar bear. They are stationary and on frictionless
level ice, initially 60 m apart.
When the hunter pulls the polar bear to
him, the polar bear will move:
1. 10 m
2. 30 m
solve A pilot heads her plane with a velocity of 250 km/h north there is of 100km/h blowing west what is the velocity of the airplane relative to the ground ?
To determine how far the polar bear will move when the hunter pulls it to him, we need to consider the conservation of momentum.
Since they are on a frictionless surface, the net external force on the system is zero (neglecting air resistance). According to Newton's second law, the force exerted by the hunter on the polar bear will be equal and opposite to the force exerted by the polar bear on the hunter.
We can use the equation for conservation of momentum to solve the problem. The equation is:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where:
m₁ and m₂ are the masses of the hunter and polar bear, respectively,
v₁ and v₂ are their initial velocities,
v₁' and v₂' are their final velocities.
In this case, since the hunter and polar bear are initially stationary, their initial velocities are zero, so the equation simplifies to:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Since the hunter and polar bear are connected by the rope, their final velocities will be the same. Let's call this common final velocity v.
Using the conservation of momentum equation, we can rewrite it as:
0 + (400 kg)(0 m/s) = (80 kg + 400 kg)v
Simplifying the equation:
0 = (480 kg)v
v = 0 m/s
Therefore, the polar bear will not move when the hunter pulls it to him. The correct answer is 0 m.